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Lecture Comments (12)

1 answer

Last reply by: Professor Selhorst-Jones
Wed May 20, 2015 11:09 AM

Post by Carolyn Miller on May 18, 2015

In Example 5 part A. wouldn't it also be possible to get the answer by converting it to exponential form(?)? Or would I not be able to use it for all problems, only for a few like the one used?

1 answer

Last reply by: Professor Selhorst-Jones
Sun Nov 9, 2014 4:36 PM

Post by Saadman Elman on November 8, 2014

A great clarification! Although example 5 didn't make sense. As you said, you are going to do logarithm equation in the next chapter.  I Will listen to it soon. Thanks!

1 answer

Last reply by: Professor Selhorst-Jones
Sun Dec 22, 2013 1:12 PM

Post by Tim Zhang on December 18, 2013

a^b=x , why "a" must greater than zero?? can't a be negative? like when a equal -5, you can still get the value of x right?

1 answer

Last reply by: Professor Selhorst-Jones
Tue Nov 19, 2013 5:47 PM

Post by Constance Kang on November 17, 2013

hey, for example 3, you said we could make u be anything right? so i made u=10 and when i put it into calculator log(42)-log(7), i end up getting 0.7781512504. What did i do wrong?

3 answers

Last reply by: Professor Selhorst-Jones
Mon Aug 12, 2013 1:40 PM

Post by Taylor Wright on July 25, 2013

Since Log(base a) of a   is equal to 1

Why wouldn't Log(base a) of a^b  equal   1^b



Therefore making  a^Log(base a) of a^b  equal to a^1^b  

Properties of Logarithms

  • We defined the idea of a logarithm in the previous lesson. A logarithm is the inverse of exponentiation:
    loga x = y     ⇔     ay = x.
    Since logarithms and exponents are so deeply connected, we might expect logarithms to have some interesting properties, just like we discovered how exponents have many interesting properties.
  • If you're interested in understanding how we figure out any of the below properties, check out the video for an explanation.
  • From the definition of a logarithm, we can immediately find two basic properties:
    loga 1 = 0                      loga a = 1
  • Logarithms and exponentiation are inverse processes: they "cancel" each other out.
    loga ax = x                      aloga x = x
  • If we take the logarithm of a power, we can "bring the power down" in front of the logarithm:
    loga xn  =  n·loga x.
  • If we have the logarithm of a product, we can split it through addition of logarithms:
    loga (M ·N)   =  loga M + loga N.
  • If we have the logarithm of a quotient, we can split it through subtraction of logarithms:
    loga
    M

    N


     



     
    =  loga M − loga N.
  • Caution! Notice that none of these properties were ever of the form loga (M+N). That's because there is just no nice formula to break apart loga (M+N).
  • If we have a logarithm that uses a different base than we want, we can change it through the change of base formula:
    logv x = logu x

    logu v


     
    .
    Notice how this allows us to change from an expression logv x to an expression that only uses logu . By using u = e or u=10, we can evaluate with any calculator.

Properties of Logarithms

Expand the below as much as possible to write it as a sum and/or difference of logarithms.
log7 (x2 ·y)
  • From the logarithm properties, we have loga (M ·N) = loga M + loga N. Thus, we can split the expression as
    log7 (x2 ·y)     =     log7 x2 + log7 y
  • We also have the property loga xn = n ·loga x, allowing us to expand it even further.
    log7 x2 + log7 y     =     2 ·log7 x + log7 y
    All the logarithms have been expanded as much as they possibly can be, so we are now finished.
2 ·log7 x + log7 y
Expand the below as much as possible to write it as a sum and/or difference of logarithms.
ln

a5
4
 
 

b3

c2+5
 


  • From the logarithm properties, we have loga (M ·N) = loga M + loga N. Thus, we can split the expression as
    ln

    a5
    4
     
     

    b3

    c2+5
     


        =     lna5 + ln
    4
     
     

    b3

    c2+5
     
  • We also have the property loga xn = n ·loga x, allowing us to expand it even further. [Remember, 4√{k} = k[1/4]]
    lna5 + ln
    4
     
     

    b3

    c2+5
     
        =     5 ·lna + 1

    4
    ·ln b3

    c2+5
  • Next, we have a property that allows us to separate fractions: loga ( [M/N] ) = loga M − loga N. This gives us
    5 ·lna + 1

    4
    ·ln b3

    c2+5
        =    5 ·lna + 1

    4
    ·
    lnb3 − ln(c2+5)
  • Remember, there is no rule to separate loga (M+N), so while it may look like we could separate ln(c2+5) further, it is actually impossible. We must leave ln(c2+5) as it is because there is no rule we can expand it further with.
  • We can do a little bit more with the other expressions, though. Begin by distributing the [1/4]:
    5 ·lna + 1

    4
    ·
    lnb3 − ln(c2+5)
        =    5 ·lna + 1

    4
    ·lnb3 1

    4
    ·ln(c2+5)
    Then use the rule about exponents one more time:
    5 ·lna + 1

    4
    ·lnb3 1

    4
    ·ln(c2+5)     =    5 ·lna + 3

    4
    ·lnb − 1

    4
    ·ln(c2+5)
    All the logarithms have been expanded as much as they possibly can be, so we are now finished.
5 ·lna + [3/4] ·lnb − [1/4] ·ln(c2+5)
Find the exact value of the below.
log4.7 4.719
  • From the properties, we know that a logarithm of a given base cancels out exponentiation of the same base. That is,
    loga ax = x
  • Thus, since we have log4.7 and an exponentiation with a base of 4.7, the two cancel each other out and all we are left with is the power:
    log4.7 4.719 = 19
19
Give an example of two numbers that show how, in general,
loga (M+N) ≠ loga M + loga N
  • To show this, we can basically pick any numbers we want for a, M, and N to show that loga (M+N) will not be equal to loga M + loga N. However, to make it easier for us to show, it would be nice to pick some numbers that simplify in a "friendly" way to make it obvious that the two sides are not equal.
  • A "friendly" set of numbers to choose is a=2 (an easy base to work with) along with M=N=4 (a number that is easy to calculate log2 of).
  • With these numbers in mind, try plugging them in for each side:
    loga (M+N)     =     log2 (4+4)     =     log2 8     =     log2 23     =     3
    Plugging in for the other side, we have:
    loga M + loga N     =     log2 4 + log2 4     =     log2 22 + log2 22     =     2+2     =     4
    Since 3 ≠ 4, we have now shown how loga (M+N) ≠ loga M + loga N is true in general.
  • Detailed Explanation: While the above is one way to show things, there are many possible ways. You can choose virtually any combination of values for a, M, and N (assuming they are all greater than 0, otherwise you will break the logarithm) to show the expression is true. In fact, the only way loga (M+N) ≠ loga M + loga N can fail to be true is if you choose your M and N very carefully (the choice of a has no effect, either way). For the non-equality to be true, the two sides must be equal:
    loga (M+N) = loga M + loga N
    However, from other logarithm properties, we know loga M + loga N = loga (M·N). So,
    loga (M+N) = loga (M·N)
    Which means that the only way for loga (M+N) ≠ loga M + loga N to not be true is if you carefully choose your M and N values such that
    M+N = M ·N
    Thus, as long as you don't carefully choose your M and N values to make the above equation true (and/or you are not extremely unlucky in your random choice of values), whatever you pick is almost guaranteed to show that loga (M+N) ≠ loga M + loga N is true in general.
There are many, many possible ways to show this. One such way is
log2 (4+4) ≠ log2 4 + log2 4
[If you're interested in a bit of discussion on the variety of possible ways to show how the expression loga (M+N) ≠ loga M + loga N is true in general, check out the last step.]
Given that log3 a = 5 and log3 b = 6, find the value of
log3 (3 ab)
  • Currently, the expression does not explicitly contain log3 a or log3 b, so we can't use either of the pieces of information given to us at the start of the question. However, if we could find a way to explicitly show those expressions inside of log3 (3 ab), then we could use them. We can work towards this by expanding the expression through the use of logarithm properties.
  • We can expand the expression using the rule loga (M ·N) = loga M + loga N:
    log3 (3 ab)     =     log3 3 + log3 a + log3 b
  • Now we clearly and explicitly have log3 a and log3 b in the expression, so we can substitute in the values we were given:
    log3 3 + 5 + 6
  • Finally, we need to figure out what the value of log3 3 is. This is pretty easy, since loga a = 1 for any a. Thus, we have
    log3 3 + 5 + 6     =     1 + 5 + 6     =     12
12
Given that ln√x = 3 and lny = 1.6, find the value of
ln
x2

y5

  • First off, it will help us to know what the values of lnx and lny are. We already know lny = 1.6, but we don't know lnx yet. To figure that out, notice that we do know ln√x = 3 and we have the rule that loga xn = n ·loga x. Thus, we can show
    ln√x = 3     ⇒     lnx[1/2] = 3     ⇒     1

    2
    ·lnx = 3     ⇒     lnx = 6
  • Currently, the expression does not explicitly contain lnx or lny, so we can't use either of those pieces of information. However, if we could find a way to explicitly show those expressions inside of what the question gave us, then we could use them. We can work towards this by expanding the expression through the use of logarithm properties.
  • We can expand the expression using the rule loga ( [M/N] ) = loga M − loga N:
    ln
    x2

    y5

        =     lnx2 − lny5
    Then, using the rule that loga xn = n ·loga x, we can show
    lnx2 − lny5     =     2 ·lnx − 5 ·lny
  • Now we clearly and explicitly have lnx and lny in the expression, so we can substitute in the values we know:
    2 ·lnx − 5 ·lny     =     2 ·6 − 5 ·1.6     =     12 − 8     =     4
4
Write the expression as a single logarithm.
3 lnu + 5 lnv
  • We want to condense the expression into a single logarithm using logarithm properties. We can begin by putting the coefficients inside the logarithms as exponents based on the property loga xn = n ·loga x:
    3 lnu + 5 lnv     =     lnu3 + lnv5
  • Next, we can combine the two logarithms with multiplication (because they're being added together) through the property loga (M ·N) = loga M + loga N:
    lnu3 + lnv5     =     ln( u3 v5 )
    At this point we're done, because everything is contained in a single logarithm. [It's important to notice that we could not have applied the properties in the opposite order. The property loga (M ·N) = loga M + loga N can only be applied when the logs do not have coefficients in front of them, so we had to move the coefficients out of the way with the other property first.]
ln( u3 v5 )
Write the expression as a single logarithm.
3 loga − 5

4
(logb − 2 logc)
  • Begin by distributing the fraction (along with the negative attached to it) so that we can clearly see how to properly apply the logarithm properties:
    3 loga − 5

    4
    (logb − 2 logc)     =     3 loga − 5

    4
    logb + 5

    2
    logc
  • We can now put the coefficients inside the logarithms as exponents based on the property loga xn = n ·loga x:
    loga3 − logb[5/4] + logc[5/2]
  • Next, we can combine the two logarithms that are adding together through the property loga (M ·N) = loga M + loga N:
    loga3 + logc[5/2]− logb[5/4]     =     log(a3 c[5/2] ) − logb[5/4]
  • Finally, we can combine the logarithm that is subtracting through the property loga ( [M/N] ) = loga M − loga N:
    log(a3 c[5/2] ) − logb[5/4]     =     log
    a3 c[5/2]

    b[5/4]

    At this point we're done, because everything is contained in a single logarithm.
log( [(a3 c[5/2])/(b[5/4])])
Evaluate the below (to three decimal places) using a calculator and the change of base formula.
log19 947
  • Most calculators only have two buttons for evaluating logarithms: ln and log (which respectively mean loge and log10). This means that if you want to evaluate a logarithm that has a different base than those two, you need a way to change the base of the logarithm. This is where the change of base formula comes in to play. [Some calculators can directly calculate the value of logarithms with bases other than e and 10. If you have such a calculator, congratulations!-You've got a real beast of a calculator. Nonetheless, play along with the below steps so that you can see how to change base in cases where the calculator can't help (like when working with variables).]
  • The change of base formula allows us to change from using the base v to the base u as below:
    logv x = logu x

    logu v
  • Thus, since we want to switch from the base of 19 to one of the bases on our calculator (e or 10-let's go with 10, just to make a choice for the below [although either is fine]), we can use the formula as follows:
    log19 947 = log10 947

    log10 19
  • Now we can use a calculator to find the value of both log10 947 and log10 19, then divide to find
    log10 947

    log10 19
    ≈ 2.328
2.328
Evaluate the below (to three decimal places) using a calculator and the change of base formula.
log√2 15
  • Most calculators only have two buttons for evaluating logarithms: ln and log (which respectively mean loge and log10). This means that if you want to evaluate a logarithm that has a different base than those two, you need a way to change the base of the logarithm. This is where the change of base formula comes in to play. [Some calculators can directly calculate the value of logarithms with bases other than e and 10. If you have such a calculator, congratulations!-You've got a real beast of a calculator. Nonetheless, play along with the below steps so that you can see how to change base in cases where the calculator can't help (like when working with variables).]
  • The change of base formula allows us to change from using the base v to the base u as below:
    logv x = logu x

    logu v
  • Thus, since we want to switch from the base of √2 to one of the bases on our calculator (e or 10-let's go with 10, just to make a choice for the below [although either is fine]), we can use the formula as follows:
    log√2 15 = log10 15

    log10 √2
  • Now we can use a calculator to find the value of both log10 15 and log10 √2, then divide to find
    log10 15

    log10 √2
    ≈ 7.814
    [If you have difficulty taking the log of √2 on your calculator (although it should work fine on the majority of calculators: just calculate √2, then take its log using the appropriate log button.), you can always use an accurate approximation of √2 in decimal form, then take its log.]
7.814

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Properties of Logarithms

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:04
  • Basic Properties 1:12
  • Inverse--log(exp) 1:43
  • A Key Idea 2:44
    • What We Get through Exponentiation
    • B Always Exists
  • Inverse--exp(log) 5:53
  • Logarithm of a Power 7:44
  • Logarithm of a Product 10:07
  • Logarithm of a Quotient 13:48
  • Caution! There Is No Rule for loga(M+N) 16:12
  • Summary of Properties 17:42
  • Change of Base--Motivation 20:17
    • No Calculator Button
    • A Specific Example
    • Simplifying
  • Change of Base--Formula 24:14
  • Example 1 25:47
  • Example 2 29:08
  • Example 3 31:14
  • Example 4 34:13

Transcription: Properties of Logarithms

Hi--welcome back to Educator.com.0000

Today, we are going to talk about the properties of logarithms.0002

In the previous lesson, we introduced the idea of a logarithm, which was defined as loga(x) = y for ay = x.0005

So, the logarithm of a number is what we would have to raise the base to, to get the number.0015

So, loga(x) = y means ay = x.0022

It is a little bit of a complex idea the first time we talk about it, so the previous lesson is really useful.0027

If you haven't already watched the previous lesson, Introduction to Logarithms, I really recommend that you watch it,0032

because that will get you a grounding in how these things work--it will really explain things, if it is confusing you.0037

Previously, when we investigated exponentiation, we found all sorts of interesting properties,0042

such as xa times xb = xa + b--that we could add exponents--0046

or that x-a = 1/xa--that we flip when we have negative exponents.0051

Since logarithms and exponents are so deeply connected (we have this idea that log of something0057

equals this other version in exponent world), we might expect logarithms to also have some interesting properties.0061

Indeed, they do: this lesson will be all about looking at the properties of logarithms.0068

Let's just start with some really basic properties--remember, this is the definition that we will be working with the whole time.0073

It is a really good idea to understand this.0079

From this, we can immediately see two basic properties: loga(1) = 0, because a0 = 1 for anything;0082

that was one of the things that we figured out when we were working with exponents;0090

and also, loga(a) is going to equal 1, because a1 is just equal to a, because it is just 1a.0093

Those are two basic properties that we get, just from the simple definition.0100

Next, we can talk about inverses: logarithms and exponentiation are inverse processes.0104

If they have the same base, they cancel each other out.0109

This is clear when a logarithm acts on exponentiation; if we have loga(ax),0114

then we are going to get x out of it, because by the definition of a logarithm,0119

the number that we need to raise x to--what do we have to raise a to, to get ax?0123

Well, that is going to be ax; so a is our base for the right side; ax = ax, so loga(ax) = x.0129

It cancels out; a logarithm on exponentiation...they cancel each other out, because we have an a here and an a here as bases.0139

So, they cancel out, and we are left with just the x that we originally had.0147

So, logarithms cancel out exponentiation, if they are of the same base.0151

We just get left with the exponent at the end.0156

To see the reverse process of exponentiation, canceling a logarithm, we first need to realize a useful idea.0158

Notice that, for any exponent base where a is greater than 0, our exponent base is positive, and it is not equal to 1.0165

And for any number x such that x is a positive number, there exists some real number b such that ab = x.0171

In other words, through exponentiation, this a to the something, we can get to any positive number x.0180

For any positive number...you say 58 as your x...then with my a, I can raise it to something that will get it to 58.0185

And then, if we want, we can name the exponent that will do this; and we called it b here.0194

There is some number b such that ab = x.0198

So, for example, we could say that 10 to the something has to be equal to 100.0202

Our question is what number it is as our something: 10 to the what equals 100?0208

Well, it is going to be 2; we raise 10 to the 2--102--we get 100.0213

So, in this case, our b, if we want to name the number in the box, was 2.0218

We could do this for something else: we could say, "3 to the box equals 47."0223

Now, this is considerably more complex than 10 to box equals 100.0230

3 isn't going to have some nice integer, some nice even power, that we can raise it to, that will put out 47.0233

But we can note that there has to be something that fits in that box.0241

We don't know what the number is right now, but we are confident--we are sure that there has to be something that we could raise 3 to, to get 47.0244

So, we can just call it b; in fact, if you were to work it out through other stuff that we will talk about as we go on in this class,0251

if you were to work it out, you would find out that b is approximately equal to 3.504555.0258

If you plug that into a calculator, you will see that it is very, very close to being precisely 47.0266

So, there is some number out there; we can't get it exactly, but we can get a very good approximation through things that we work out in this course.0270

We can be sure that this number does exist--it is out there somewhere, even if we don't know it.0278

But that doesn't mean we can't talk about it; we can not know what something is precisely, and still talk about it as a general idea.0283

So, we know that this has to be true, because every exponential function, a to the something, has a range of 0 to infinity.0291

We have talked about this before, and you can see it in the graphs of it.0297

Since it has a range of 0 to infinity--since any number can come out of it, as long as it is a positive number--0300

then there has to be some b that will make that number from between 0 and infinity.0305

For any x contained in 0 to infinity, for any positive number x, there has to be some b that,0310

when we plug it in here, we will end up getting ab = x.0315

We will end up getting that x when we plug it in.0322

This idea that there is always an exponent for creating any positive number (it won't work if it is negative;0324

but for creating any number) is really important; we will use this fact to prove a variety of properties about logarithms.0329

So, we want to keep this idea that, if we have some x, and we know we have some base a,0335

then there has to exist some real number b to make ab = x.0340

So, you start with a base a; you start with some sort of destination x; you are guaranteed that there is a b that will get you to that destination through exponentiation.0345

With this idea in mind, we can now show inverses in the other direction.0354

Consider aloga(x): now, by our new idea, we know that there has to be some b such that ab = x.0359

Here is our x; and we just choose the base a--it is going to be connected to the fact that we have the base a here, showing up here.0368

So, we know that a to the something equals x; and we are guaranteed that there has to be some b there.0375

So, we are talking about it here: we know that there is some b such that ab = x.0381

So, we can say a to the loga of...and now we replace it with ab = x,0385

so we replace right here, and it becomes ab, so aloga(ab).0392

Now, we just showed that loga(ab) = b, because logs on exponents cancel out.0399

So, loga(ab) = b, so we have the same thing here.0406

loga(a) will drop us down to just having the b, so that b will move over to being the exponent, and we will have ab.0410

But how did we define ab in the first place? It was ab = x.0418

So, we know that ab = x; since we created b based on the fact that ab = x,0422

we have now shown that what we started with ends up being equivalent to x.0428

So, we have that a raised to the loga(x) is equal to x.0432

So now, at this point, we have shown inverses in both directions.0436

If you take log of an exponent, and they both have the same base, they cancel out, and we are just left with whatever was our power.0439

If we take some base and raise it, through exponentiation, to a log, and that base there and the base on the log are the same thing,0445

we end up just having what the log was taking as its quantity.0455

So, we have inverses shown in both ways; we can cancel out in both directions.0460

All right, we will move on to something else: the logarithm of a power.0465

Consider if we had some positive number x, and any real number n; then what if we raised to the n and took its log?0468

So, we have loga(xn); by our key idea,0473

we know that there exists some b such that we can write ab = x--the same thing.0478

So, if we want, we can swap out our x here for ab.0484

So, we have ab to the n now; from our rules about exponents, we know that ab to the n,0488

is the same thing as just a to the b times n; so now, we have b times n, or we could write it as n times b.0499

loga(a) to the n times b becomes canceled out, because we have inverses there,0506

since they are the same base, and we are left with n times b.0514

Now, do we have another way to express b?0517

Well, from the beginning, we know that ab = x, so we could write that in its logarithmic form.0519

And we would show that loga(x) = b, because ab = x.0525

If we move this over here, a raised to the b becomes x from how we had this originally.0534

So, we see that loga(x) = b, which means that we can swap the b out here for the loga(x).0539

And we have that loga(x) here; we started with loga(xn),0548

and now we see that we can take this n here and move it out front, and we have n times loga(x).0555

So, if we have an exponent, we can move it out front.0562

Let's look at an example: if we look at log3(32), by this property,0565

we see that we could take this 2 and move it out front, and we would have 2 times log3(3).0574

Well, log3(3)...log base something of something, if they are the same something...is just 1.0580

What do you have to raise 3 to, to get 3? Just 1.0587

So, we have 2 times 1, which is equal to 2.0590

What if we look at it the other way?0595

Well, log3(9)...what do we have to raise 3 to, to get 9?0596

We just have to raise it to 2; so we have 2 here and 2 here; either way we go at it, we end up getting the same thing.0600

We see this property in action.0606

OK, now let's consider two positive numbers, m and n (that should be "numbers, m and n"): loga(mn).0609

log base a of the whole quantity, m times n: now, from our key idea, we know that there exist m and n,0618

such that we can raise a to the m, and will get M; and we raise a to the n, and we get N.0624

So, we can swap these things out: we can swap out here, and we can swap out here.0631

And we end up getting loga(aman).0636

Now, from our work about exponents, we know that am times an is the same thing as am + n.0642

The same base means that we can add the exponents, so we have loga(am + n).0649

And then, since it is loga on an exponent base of a, we get cancellation once again through inverses;0655

loga cancels out with that, and we are left with just m + n.0661

Now, we can ask ourselves, "Do we have another way to express m and n?"0665

Well, from the beginning, we know that am = M; that was how we set this up.0668

So, loga(M) = m, because we know that if we raise a to the m, we get M; so we have that.0677

We can express it in its exponential form or its logarithmic form.0691

So we now can say, "What about looking at it through its logarithmic form?"0694

The same thing goes over here with an = N; we can express it, instead, as loga(N) = n.0697

So, at this point, we have two different new ways to be able to describe m + n.0705

We can swap that out; and so, loga(M) here becomes here, and loga(N) = n goes here.0712

So, you see this here; you see this here; we can swap that out.0723

So, what we started with, loga(MN), is going to be the same thing as loga(M) + loga(N).0729

So, if we have a logarithm of a product, we can split it into the sum of the logarithms of the numbers that made up each part of that product.0741

Let's look at an example to help clarify this.0753

We could look at log10; we will write it as just log, since the common log can be expressed as just log.0756

So, log(1000): we could write this, if we felt like it, as log(100 times 10).0768

And then, by this rule that we have here, we could split it; we have products 100 and 10, so we can split it into log(100) + log(10).0777

Now, what number do we have to raise 10 to, to get 100?0786

We have to raise it to 2; what number do we have to raise 10 to, to get 10? We only have to raise it to 1.0789

So, 2 + 1...we end up getting 3.0801

Alternatively, we could have done this as log(1000): what number do we have to raise 10 to, to get 1000?0804

10 is 1; 100 is 2; 1000 is at 3; so we could also see that log10(1000) = 3 over here.0810

So, it works out either way that we want to approach it.0820

So, we see now that we can break up products into two different things being added together.0822

What if we took the logarithm of a quotient, loga(M/N)?0829

Well, we could rewrite that as M times N-1, since we could rewrite M/N as M times 1/N.0833

And then, we can rewrite 1/N as just its flip, so we would have N-1.0843

We can swap it around like that.0848

From the rule that we just saw, the splitting of products, we can write this as...0850

here is the M, so we have loga(M); and here is the N-1, so we have loga(N-1).0855

So, loga(M)...and then, we also have our rule that we can bring down exponents.0861

log(xn) becomes nlog(x); so we bring this down in the front.0867

Since we are bringing down a -1, it just becomes a minus sign here; so we have loga(M) - loga(N).0873

Thus, loga(M/N) is equal to loga(M) - loga(N).0884

So, the logarithm of a quotient becomes the difference of the two logarithms.0889

Let's look at an example to help clarify this one, as well.0896

So, if we look at log2(32/2), we could write that, by our new rules, as log2(32)0899

(is the part on the top), and then the part on the bottom is 2, so minus log2(2).0911

log2(32): what number do we have to raise 2 to, to get 32?0920

Well, 2 is 1; 4 is 2; 8 is 3; 16 is 4; 32 is at 5--so we raise 2 to the 5, and we get 32; so log2(32) is 5.0924

What is log2(2)? -1: 2 to the 1 equals 2, so just 1; so 5 - 1...we end up getting 4.0937

What if we had looked at it, not through breaking it apart, but just simplified it first--32/2?0948

Well, we could write that as log2(16); what number do we need to raise 2 to, to get 16?0953

2 to the 1 becomes 2; 2 squared becomes 4; 2 to the 3 becomes 8; 2 to the 4 becomes 16; so this would end up being 4.0959

So, we end up getting the same thing, either way we look at it.0968

Great; I want to caution you that there is no rule for log(M + N).0971

Notice: none of these properties were ever of the form log(M + N) inside of there.0979

That is because there is just no nice formula to break apart log(M + N).0984

So, if you have a log of a quantity, and inside of that quantity it is something plus something else, there are no special rules.0989

Sorry--there is just no easy way around it; you are going to have to work things out there in a complicated way.0994

There is no way to be able to just break things apart or put things together anymore.0998

Lots of people make the mistake of thinking that loga(M + N) is equal to loga(M) + loga(N).1002

or that loga(M - N) is equal to loga(M) - loga(N).1007

Those are not true--not true at all; you can't write them and split them apart like this.1011

These things do not work; it is the same thing as, if you have √(2 + 2), saying, "Oh, I will just split that into √2 + √2."1017

That does not work; you can't just split on square roots; you can't just split on logarithms, either.1027

So, this idea of just splitting because you see an addition sign--you can't do that.1032

You have to work out what is the logarithm of everything inside of there; there are no clean rules to do that.1036

It is an easy mistake to end up making; but don't let it happen to you.1041

Be vigilant; watch out for this; don't let it happen to you; don't do the same mistake.1044

Remember, it only works with M times N and M divided by N.1048

If it is plus or minus, there are no special rules; you just have to work it out by figuring things out,1052

simplifying, hopefully, if you can, like they are actually numbers; but there is really no easy way around it.1056

At this point, we have seen a lot of properties for logarithms; so let's review them.1063

Our base ones right at the beginning were the log base any a of 1 equals 0,1065

because any a raised to the 0 becomes 1; and also, log base a of a equals 1, because any a raised to the 1 is just a.1071

Then, we also have our inverse properties, loga(ax) = x,1079

that it cancels out when you have logarithm on exponentiation if they are the same base;1083

and then a raised to the loga(x) = x when we have exponentiation acting on logarithms.1088

It cancels out if they are the same base.1092

Then, we have the fact that we can bring down powers.1095

If we have log(xn), then we can bring the n in front: we have n times log(x).1098

If we have log(MN), then we can split that into log(M) + log(N).1105

We have the two different pieces, M and N, so it splits into log(M) and log(N).1111

Multiplication inside of a logarithm becomes addition outside of the logarithm.1116

loga(M/N) is log(M) - log(N); if we have M here and N here, then we have this minus sign right here.1121

So, division inside of the logarithm becomes subtraction outside of the logarithm, once we split it into two logs.1133

So, it seems like a lot of rules, and there are a fair bit of new things that you have to get used to here.1141

But they are all based off of our original definition of what it means to be a log--the idea that loga(x) = y means that ay = x.1145

This is really what it is: you can either write it in exponential form or logarithmic form.1153

It is just a way of denoting things.1157

So, a underneath that y, ay, becomes what we were originally taking the log of.1158

And so, for any base a, we also figured out this key idea that for any base a, and any positive number x,1165

there was a b that allowed us to get to that x: ab = x, for any x that we wanted to get to.1171

So, these two ideas--you can put them together, and you can figure out pretty much any one of these things right here with those ideas.1177

And then, these ones are all just coming off of the basic definition, right from the beginning.1186

But if you take that key idea, as well, you can figure these out.1192

So, if you ever forget them on a test, in a situation where you can't just look them up,1194

you now have a way of hopefully being able to figure them out on your own.1197

They are not quite as easy as being able to figure out all of the things that made up our exponential rules,1200

our rules for exponentiation, but we are able to figure these things out on our own.1204

And as you get more used to working with them, and get some practice in them,1209

it will be even easier for you to work them out on your own.1212

And they will also just stick in your head that much easier.1214

All right, now we are going to switch to a new idea.1217

In the last lesson, we mentioned that most calculators only have buttons to evaluate natural log of x and log(x),1220

ln(x) and log(x), that is log base e (that is what natural log means) and log without a number (means base 10).1225

So, how could we evaluate something like log7(42)?1234

Well, 7 to the 1 equals 7, and 7 squared equals 49; so we see that there is no easy integer number that we raise 7 to, to get 42.1237

It is not going to be an easy thing, so we need to use a calculator, because 42 is not an integer power of 7.1248

But since it is base 7, we don't have a button on our calculators.1254

What we want is some way to transform the base of the logarithm.1257

If we could transform from log base 7 into log base 10 or log base e, we would be able to use a calculator,1260

because then we have our natural log and common log buttons on our calculator; we can just punch it in.1266

Now, you might have a calculator that lets you just put in an expression like this.1271

But even if you have that sort of calculator, this is still sort of a useful thing to learn.1275

As we will see in some of the examples, there are ways to apply changes of base beyond just using them to get what these numbers are.1279

And also, lots of times, you won't have a calculator that is able to do this, and you will only have natural log or plain common log, log base 10.1285

And you will need to be able to have this change of base, so that you can change when you need to take a base that isn't e or 10.1293

To help motivate the coming formula and its derivation, let's look at a specific example.1300

Consider the expressions log3(81) and log9(81).1304

log3(81): well, we could rewrite 81 as 34, so we see that that is just 4 over here.1308

Now, log9(81) we could rewrite, also, as 92; so with a base 9, that would come out as 2.1315

So, we see that log3(81) = 2log9(81), because 4 is equal to 2 times 2.1323

So, we see that log3(81) is equal to 2 times log9(81).1333

And this is somehow related to the fact that 32 is equal to 9.1341

You square 3, and you manage to get a 9 out of it.1345

Now, we can probably intuitively know that that holds, generally.1348

We can figure out that normally (or in fact, always), we are going to have log3(x) = 2log9(x).1350

But let's prove it, instead of just assuming--instead of getting a feel that that makes sense, let's actually prove definitely that that is the case.1358

We start by noting that x is equal to 9 to the log9(x).1365

Remember: if we wanted to, we could cancel those things out, because we have a base of 9 and log base 0.1369

So, they would cancel out, and we would end up having just x; so this here makes sense.1378

But having it, 9 to the log9(x), written in this funny way, is a complicated idea.1382

It is hard to see where we pulled it; it is kind of like just pulling a rabbit out of a hat.1386

But with this idea, if we leave it in this form, we will be able to do some cool tricks that will let us show what we want to get to.1390

If we want, we can take log3 of both sides.1396

We know that x is equal to 9log9(x), so it must be the case that log3(x) is the same thing as 9log9(x).1398

So, log3(9log9(x)).1410

We can take log base 3 of both of the things on either side, because that equals sign means that it has to be equal for whatever happens to it.1413

So, log3(x) is equal to log3(9log9(x)).1419

OK; with that idea in mind, we can start applying our rules that we have.1423

We know that we can bring down exponents; so in this case, we have effectively an exponent of log9(x).1427

We actually have it exactly as an exponent, so if we want, we can bring that down in front.1433

We see that log3(x) is equal to log9(x) times log3(9).1437

Now, what is log3(9)? That comes out to be 2, so we can simplify that as log3(x) = 2log9(x).1442

And we have proven what we originally wanted to show.1452

We follow a similar structure to create a formula to change between any two bases, u and v.1456

If we start as x = vlogv(x), which we know is true, because of inverses,1460

then we can once again take a log on both sides; and we will take logu, because we want to get v and u.1465

We want to get both of those logs into action, so we take logu on both sides.1471

And then, we can bring this down in front, because it is an exponent; and we have logu(x) = logv(x) times logu(v).1475

At this point, we can create a formula that will have logv on one side and logu on the other side.1483

We divide logu(v) over, and we rearrange; we swap the order of the equation.1488

We have logv(x) = logu(x)/logu(v).1493

Notice that this allows us to change from an expression logv(x) into an expression that only uses log base u.1503

Now, if we choose our u to be either e or 10 or whatever is convenient for the problem we are working on,1510

we will be able to evaluate it with any calculator at all.1515

Since every calculator we will be using has natural log and common log (base 10 log) buttons,1518

we will be able to evaluate with any calculator, because we will be able to change the u's over here.1523

So, whatever we end up having--if it is log7(42), then we can change it into log10(42)/log10(42).1528

Or alternatively, we could have changed it into ln(42)/ln(42).1538

Both would end up giving the same thing; and we will see what that is in the examples.1544

All right, the first one: Write as a sum and/or difference of logarithms.1549

Our first example here: we are working with base 5, but that doesn't affect how any of our properties work.1552

So, remember: if we have log(M/N), for any base a, then that is equal to loga(M) - loga(N).1558

The same log base is on both, and it splits into subtraction.1570

So, in this case, we have, on the top, x5; so we will have log5, the same thing,1573

minus what is on the bottom, y√z, so - log5(y√z).1580

Great; now, we also have the rule that loga, for any a of M times N, equals loga(M) + loga(N).1596

So, we can split with addition, as well; so we have multiplication here.1609

y times √z is what is really there.1613

log5(x5) -...now, notice: we are splitting all of this here.1617

We are still subtracting by all of it, so we want to put parentheses around it, because it is substitution that we are doing here.1625

So, we now work on this thing here, loga(M) - loga(N).1632

So, our M is y; our N is √z; and we have log...still the same base...of y + log, still the same base (5), of √z.1637

Simplify this out a bit: we have log5(x5) - our subtraction distributes...minus here, as well... log5(√z).1648

Now, that is technically enough, because we have a sum and/or difference of logarithms.1662

But we can also take it one step further, and we can get rid of these exponents.1665

We can get rid of "to the fifth"; we can get rid of √z, because we also have the rule that loga,1668

for any base, of xn, is equal to n times loga(x).1674

So, in this case, we have to the fifth; and how can we rewrite √z?1682

Well, remember: any square root is just like raising to the half, so we can see this:1686

bring the 5 to the front, so we have 5log5(x).1695

We continue to just bring down our log5(y) -...we will rewrite the z...log5(z1/2).1701

And now, we can bring down this, as well...not to there, but we have to move it all the way to the front of the log.1711

There we go: we have 5log5(x) - log5(y) - 1/2log5(z).1718

And there we go: we have managed to write this entirely using very simple things inside of our log: just x, y, and z.1732

We have managed to break up this fairly complicated expression inside of the log into a fairly simple expression1739

inside of the log by just breaking it up into more arithmetic.1745

We can do the reverse, and we can also compact things: write the expression as a single logarithm.1749

We will compact all of these log expressions into one tiny log with a more complicated structure on the inside.1753

So, first, we have, once again, that subtraction becomes division: so 1/3ln(a) + 2...1761

actually, the first thing: we have these coefficients out front.1769

We can also bring the coefficients in, so 1/3 can hop up onto an exponent on that a.1774

The 2 here can hop up to an exponent on that b, so we have ln(a3) + 2(ln(b2)) - ln(c).1780

If you forgot, remember: natural log, ln, is just a way of saying log base e, where e is a special number, the natural base.1794

ln(a1/3) + 2(ln(b2)) - ln(c).1801

Now, we can compact what is inside of those parentheses, because we see we have subtraction.1805

So, that is natural log of b2/c; subtraction of logs is the same thing as division inside of the logarithm.1812

And now, natural log a1/3 plus...well, now we have this 2, so we can take this 2,1821

and it is hitting a log; it is times a log, so it can go up and also become an exponent.1829

So, ln(b2/c)...make sure we remember that we are doing the whole logarithm of that whole thing,1834

so ln(a1/3) + ln...we distribute that...(b4/c2).1844

And we can also bring this in now: natural log of...addition of logarithms becomes multiplication inside of the logarithms.1853

So, a1/3 times the rest of it...b4...it will show up in the numerator, divided by c2.1860

And we have the whole thing compacted into a single logarithm; great.1868

All right, the next one: Evaluate each of the following; use a calculator and the change of base formula.1873

Remember our change of base formula: if we have log7(42), then we can change to any base...1880

I'll put it as a square right now...of 42; the thing that we are taking our log of initially, divided by...1888

the base has to be the same between here and here; these have to be the same base.1899

And it is going to be of our original base; so since it was log7(42), we now have log of something (42), divided by log of something (7).1905

All right, that is how it works: that is what it meant when we saw logv(x) = logu(x)/logu(v).1915

In this case, for this one, our v is 7; our u is whatever we are about to choose.1927

So, we can make it any u we want; we can make it 50, and it would work.1935

We could make it .1, and it would work; but let's do something that shows up on our calculators.1938

So, let's choose e: we can put in an e here and an e here.1942

So, we will rewrite that as ln(42)/ln(7); we punch that into our calculator, and that will end up coming out to be...1948

it will go on with lots of decimals, so let's cut it off, and it will end up being approximately 1.9208.1957

Now, if you wanted to, you could have also done this as something else.1965

It would have also been the same as log10(42)/log10(7).1968

That would be the same if you used common log, something that also shows up on a lot of calculators.1976

And you would end up getting the same thing; it would come out to be approximately 1.9208.1981

Now, if we want to check our work--we want to make sure that that did work out--1986

we would check it by saying that 71.9208 does come out to being 42.1990

And it does come out to being approximately 42, so it ends up checking out if we punch that into a calculator.1998

Let's do logπ(√17); it is the same basic idea here.2004

We can do it as ln(√17)/ln(π); we punch that out in a calculator, and we get approximately equal to 1.2375.2007

And if we wanted to, we also could have done that as log10(√17)/log10(π).2022

And we would have ended up getting the exact same thing.2029

It would have come out to be approximately 1.2375.2032

Either one you work with--they are both going to end up working out to give you the same answer.2034

Great; and if you wanted to, you could also check this, as well.2038

You could check and make sure that π raised to approximately 1.2375 does come out to be approximately √17.2041

And indeed it does, if you want to check that this does work.2049

Great; the fourth example: Given that log5(a) = 6, and log5(b) = 1.2, evaluate each of the following.2053

At this point, we want to use the rules that we have to split things apart.2061

Splitting it apart will allow us to use the pieces of information we have.2064

We have to see a log5(a) before we can swap it out for 6; so we have to get that sort of thing to show up--the same for log5(b).2068

So, we split things up; we have multiplication between each one of these--that is what it means when they are just stacked on top of each other.2075

So, log5(5) + log5(a) + log5(b3):2084

log5(5) is just 1, because it is the same thing as its base, and what it is operating on,2095

plus log5(a); we were told that was 6, so we get 6;2100

plus...now, log5(b3)...we can't do that yet.2104

We need to get it as just a b inside; but we see that there is an exponent.2107

We can bring that out front; so we have 3log5(b).2112

So now, we can use the fact that it is 1.2: 1 + 6 is 7, plus 3 times 1.2 (is 3.6); 7 + 3.6 becomes 10.6; and there is our answer.2117

Work on the other one: log25(√a); this one is a kind of a problem.2139

We have 25 here, but we were told our base for working with the stuff--the information we were given was log base 5.2144

So, we have to use change of base; now we see a time when you have to use change of base,2149

not just for calculating numbers (if you have a calculator that can do change of base on its own,2153

that doesn't need you to do it, then there is still a use for it for problems like this).2158

Change of base: we have that this log25(√a) is the same thing2162

as log of same base of √a, divided by log of some base of what our original base was, 25.2168

So, what do we want to use there? We probably want to use 5, because that is the thing we have all of our information on.2179

So, 5 and 5 here; log5(√a)...now, √a is not what we have--we have a.2184

Is there another way to write √a that would involve a?2193

Yes, √a is the same thing as a1/2: so we can write that as log5(a1/2).2196

divided by log5(25)...well, we see that that is log5(52),2203

because what number do you have to raise 5 to, to get 25? You have to raise it to 2.2209

So at this point, 1/2log5(a)/2log5(5)...log5(5) is just going to be 1.2213

You swap out; we know that we have 6 up here, so it is 1/2 times 6, divided by 2; 1/2 times 6 is equal to 3, still divided by 2.2226

There we go--cool.2239

All right, the final example: Solve the following equations.2241

We didn't talk about how to solve equations like this very much in detail, because hopefully this idea will make sense.2244

But don't worry; we are going to talk about this in great detail in the next lesson, where we will really get into this.2250

But we are just going to start with some simple ones, just in case you end up having any problems like this already that you are working on.2255

So, solve the following equations: log3(x + 5) = 2.2260

Now, remember: we have that inverse property--we know that a to the loga of "something"2264

ends up being equal to something, because these cancel out.2272

Now, notice: we have an equals sign here, so we know that log3(x + 5) is the same thing as 2.2276

So, we can use either one either way we want to.2282

So, that means that a to the something equals a to the something, if it is the same something.2285

So, why don't we choose 3 as our a? 3 to the stuff is equal to 3 to the stuff, as long as it is the same stuff.2290

Let's put 2 over here and log3(x + 5) over here.2301

32 and 3log3(x + 5): 3 and log3 end up canceling out,2309

and x + 5 just drops down; that equals...there is nothing to cancel on the right side; it is 32;2325

but we know what 32 is: 3 times 3 is 9.2331

x + 5 = 9: we subtract 5 on both sides, and we get x = 4.2333

Great; and if we wanted to, we could check that that does end up working out.2338

Check: we plug in our x = 4, so log3(4 + 5) = 2; does it?2342

log3(4 + 5)--let's see if that ends up coming out to be 2.2352

log3(9): what number do we have to raise 3 to, to get 9?2357

We have to raise it to 2; so it checks out.2361

Great; ex - 8 = 47: to do this one, we remember that loga(ax) = x.2364

A log on an exponent, as long as they are both the same base, also cancels out.2375

So, what is the base for e? it is natural log, ln; we could also do loge, but normally it is done as ln.2381

We can take the natural log of both sides; just as we had 3 to the something equals 3 to the something,2388

the natural log of something equals the natural log of something, as long as it is the same something.2393

So, ln(stuff) is equal to ln(stuff), as long as it is the same stuff.2397

Well, we have this right here; so we know that we can plug in 47 over here, and ex - 8 we can plug in over here,2404

because we are guaranteed by that equals sign that it is the same stuff on either side, that they end up being the same thing.2412

So, ln(ex - 8)--well, natural log is just log base e, so these cancel out; and the x - 8 drops down.2417

And that equals ln(47)...well, that is going to end up coming out to be a pretty not-simple number.2425

It is going to have a lot of decimals; so let's just leave it as ln(47) for right now.2431

And we end up getting, by adding 8 to both sides, ln(47) + 8.2435

Now, alternatively, we could also figure out what this is as a decimal approximation.2446

So, if we punch ln(47) into our calculator, and then add 8 to it, we get approximately 11.85.2450

That would also be approximately 11.85.2459

It is precisely ln(47) + 8, but if we want to take ln(47) and have a number that we can work with, it ends up coming out to something with a lot of decimals.2461

It is just like when you have 2 times π you can leave the answer precisely as 2π.2470

But if you want to, you can also approximate that into 6.28...and there is more stuff to it.2477

So, 2 times π is the exact answer; but you also might want a decimal answer to work with, so you can approximate it by multiplying it out.2485

The natural log of 47 is the same sort of thing as in the π example.2491

It is something that is a complicated number, so we might want to leave it precisely, or we might want to get it approximately.2495

We can check with this number; and we have that e to the...let's use our decimal approximation,2500

so we can actually put it into a calculator...11.85, minus 8...what do we put that in?2505

That will become e3.85; e3.85 ends up being approximately 46.99.2511

So, that ends up checking out, because ultimately, remember, we just said it was an approximation, not perfectly the answer.2520

The thing that is perfectly the answer is this one right here.2526

This is pretty great stuff that allows us to get a whole bunch of applications worked out with this,2529

as we will see two lessons from now, when we talk about applications of the stuff.2532

And in the next lesson, we will really dive into how we solve equations like this.2536

So, if you want more information on that, check out the next lesson,2540

where we will really see some really complicated examples, and get a really great idea of how these things work.2543

And we will really understand how to solve all of these sorts of equations.2548

All right, we will see you at Educator.com later--goodbye!2551