For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Properties of Logarithms
 We defined the idea of a logarithm in the previous lesson. A logarithm is the inverse of exponentiation:
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.log_{a} x = y ⇔ a^{y} = x.  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:
log_{a} 1 = 0 log_{a} a = 1  Logarithms and exponentiation are inverse processes: they "cancel" each other out.
log_{a} a^{x} = x a^{loga x} = x  If we take the logarithm of a power, we can "bring the power down" in front of the logarithm:
log_{a} x^{n} = n·log_{a} x.  If we have the logarithm of a product, we can split it through addition of logarithms:
log_{a} (M ·N) = log_{a} M + log_{a} N.  If we have the logarithm of a quotient, we can split it through subtraction of logarithms:
log_{a} ⎛
⎝M N
⎞
⎠
= log_{a} M − log_{a} N.  Caution! Notice that none of these properties were ever of the form log_{a} (M+N). That's because there is just no nice formula to break apart log_{a} (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:
Notice how this allows us to change from an expression log_{v} x to an expression that only uses log_{u} . By using u = e or u=10, we can evaluate with any calculator.log_{v} x = log_{u} x log_{u} v
.
Properties of Logarithms

 From the logarithm properties, we have log_{a} (M ·N) = log_{a} M + log_{a} N. Thus, we can split the expression as
log_{7} (x^{2} ·y) = log_{7} x^{2} + log_{7} y  We also have the property log_{a} x^{n} = n ·log_{a} x, allowing us to expand it even further.
All the logarithms have been expanded as much as they possibly can be, so we are now finished.log_{7} x^{2} + log_{7} y = 2 ·log_{7} x + log_{7} y

 From the logarithm properties, we have log_{a} (M ·N) = log_{a} M + log_{a} N. Thus, we can split the expression as
ln ⎛
⎜
⎝a^{5} 4⎛
√b^{3} c^{2}+5⎞
⎟
⎠= lna^{5} + ln 4⎛
√b^{3} c^{2}+5  We also have the property log_{a} x^{n} = n ·log_{a} x, allowing us to expand it even further. [Remember, ^{4}√{k} = k^{[1/4]}]
lna^{5} + ln 4⎛
√b^{3} c^{2}+5= 5 ·lna + 1 4·ln b^{3} c^{2}+5  Next, we have a property that allows us to separate fractions: log_{a} ( [M/N] ) = log_{a} M − log_{a} N. This gives us
5 ·lna + 1 4·ln b^{3} c^{2}+5= 5 ·lna + 1 4· ⎛
⎝lnb^{3} − ln(c^{2}+5) ⎞
⎠  Remember, there is no rule to separate log_{a} (M+N), so while it may look like we could separate ln(c^{2}+5) further, it is actually impossible. We must leave ln(c^{2}+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]:
Then use the rule about exponents one more time:5 ·lna + 1 4· ⎛
⎝lnb^{3} − ln(c^{2}+5) ⎞
⎠= 5 ·lna + 1 4·lnb^{3} − 1 4·ln(c^{2}+5)
All the logarithms have been expanded as much as they possibly can be, so we are now finished.5 ·lna + 1 4·lnb^{3} − 1 4·ln(c^{2}+5) = 5 ·lna + 3 4·lnb − 1 4·ln(c^{2}+5)

 From the properties, we know that a logarithm of a given base cancels out exponentiation of the same base. That is,
log_{a} a^{x} = x  Thus, since we have log_{4.7} and an exponentiation with a base of 4.7, the two cancel each other out and all we are left with is the power:
log_{4.7} 4.7^{19} = 19

 To show this, we can basically pick any numbers we want for a, M, and N to show that log_{a} (M+N) will not be equal to log_{a} M + log_{a} 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 log_{2} of).
 With these numbers in mind, try plugging them in for each side:
Plugging in for the other side, we have:log_{a} (M+N) = log_{2} (4+4) = log_{2} 8 = log_{2} 2^{3} = 3
Since 3 ≠ 4, we have now shown how log_{a} (M+N) ≠ log_{a} M + log_{a} N is true in general.log_{a} M + log_{a} N = log_{2} 4 + log_{2} 4 = log_{2} 2^{2} + log_{2} 2^{2} = 2+2 = 4  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 log_{a} (M+N) ≠ log_{a} M + log_{a} 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 nonequality to be true, the two sides must be equal:
However, from other logarithm properties, we know log_{a} M + log_{a} N = log_{a} (M·N). So,log_{a} (M+N) = log_{a} M + log_{a} N
Which means that the only way for log_{a} (M+N) ≠ log_{a} M + log_{a} N to not be true is if you carefully choose your M and N values such thatlog_{a} (M+N) = log_{a} (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 log_{a} (M+N) ≠ log_{a} M + log_{a} N is true in general.M+N = M ·N


 Currently, the expression does not explicitly contain log_{3} a or log_{3} 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 log_{3} (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 log_{a} (M ·N) = log_{a} M + log_{a} N:
log_{3} (3 ab) = log_{3} 3 + log_{3} a + log_{3} b  Now we clearly and explicitly have log_{3} a and log_{3} b in the expression, so we can substitute in the values we were given:
log_{3} 3 + 5 + 6  Finally, we need to figure out what the value of log_{3} 3 is. This is pretty easy, since log_{a} a = 1 for any a. Thus, we have
log_{3} 3 + 5 + 6 = 1 + 5 + 6 = 12

 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 log_{a} x^{n} = n ·log_{a} 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 log_{a} ( [M/N] ) = log_{a} M − log_{a} N:
Then, using the rule that log_{a} x^{n} = n ·log_{a} x, we can showln ⎛
⎝x^{2} y^{5}⎞
⎠= lnx^{2} − lny^{5} lnx^{2} − lny^{5} = 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

 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 log_{a} x^{n} = n ·log_{a} x:
3 lnu + 5 lnv = lnu^{3} + lnv^{5}  Next, we can combine the two logarithms with multiplication (because they're being added together) through the property log_{a} (M ·N) = log_{a} M + log_{a} N:
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 log_{a} (M ·N) = log_{a} M + log_{a} 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.]lnu^{3} + lnv^{5} = ln( u^{3} v^{5} )

 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 4logb + 5 2logc  We can now put the coefficients inside the logarithms as exponents based on the property log_{a} x^{n} = n ·log_{a} x:
loga^{3} − logb^{[5/4]} + logc^{[5/2]}  Next, we can combine the two logarithms that are adding together through the property log_{a} (M ·N) = log_{a} M + log_{a} N:
loga^{3} + logc^{[5/2]}− logb^{[5/4]} = log(a^{3} c^{[5/2]} ) − logb^{[5/4]}  Finally, we can combine the logarithm that is subtracting through the property log_{a} ( [M/N] ) = log_{a} M − log_{a} N:
At this point we're done, because everything is contained in a single logarithm.log(a^{3} c^{[5/2]} ) − logb^{[5/4]} = log ⎛
⎝a^{3} c^{[5/2]} b^{[5/4]}⎞
⎠

 Most calculators only have two buttons for evaluating logarithms: ln and log (which respectively mean log_{e} and log_{10}). 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:
log_{v} x = log_{u} x log_{u} v  Thus, since we want to switch from the base of 19 to one of the bases on our calculator (e or 10let's go with 10, just to make a choice for the below [although either is fine]), we can use the formula as follows:
log_{19} 947 = log_{10} 947 log_{10} 19  Now we can use a calculator to find the value of both log_{10} 947 and log_{10} 19, then divide to find
log_{10} 947 log_{10} 19≈ 2.328

 Most calculators only have two buttons for evaluating logarithms: ln and log (which respectively mean log_{e} and log_{10}). 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:
log_{v} x = log_{u} x log_{u} v  Thus, since we want to switch from the base of √2 to one of the bases on our calculator (e or 10let'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 = log_{10} 15 log_{10} √2  Now we can use a calculator to find the value of both log_{10} 15 and log_{10} √2, then divide to find
[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.]log_{10} 15 log_{10} √2≈ 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 screencaptured 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
 Inverselog(exp) 1:43
 A Key Idea 2:44
 What We Get through Exponentiation
 B Always Exists
 Inverseexp(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 BaseMotivation 20:17
 No Calculator Button
 A Specific Example
 Simplifying
 Change of BaseFormula 24:14
 Example 1 25:47
 Example 2 29:08
 Example 3 31:14
 Example 4 34:13
Precalculus with Limits Online Course
Transcription: Properties of Logarithms
Hiwelcome 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 log_{a}(x) = y for a^{y} = x.0005
So, the logarithm of a number is what we would have to raise the base to, to get the number.0015
So, log_{a}(x) = y means a^{y} = 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 workit 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 x^{a} times x^{b} = x^{a + b}that we could add exponents0046
or that x^{a} = 1/x^{a}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 propertiesremember, 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: log_{a}(1) = 0, because a^{0} = 1 for anything;0082
that was one of the things that we figured out when we were working with exponents;0090
and also, log_{a}(a) is going to equal 1, because a^{1} 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 log_{a}(a^{x}),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 towhat do we have to raise a to, to get a^{x}?0123
Well, that is going to be a^{x}; so a is our base for the right side; a^{x} = a^{x}, so log_{a}(a^{x}) = 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 a^{b} = 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 a^{b} = 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 210^{2}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 confidentwe 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 existit 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 infinitysince any number can come out of it, as long as it is a positive number0300
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 a^{b} = 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 a^{b} = 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 a^{loga(x)}: now, by our new idea, we know that there has to be some b such that a^{b} = x.0359
Here is our x; and we just choose the base ait 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 a^{b} = x.0381
So, we can say a to the log_{a} of...and now we replace it with a^{b} = x,0385
so we replace right here, and it becomes a^{b}, so a^{loga(ab)}.0392
Now, we just showed that log_{a}(a^{b}) = b, because logs on exponents cancel out.0399
So, log_{a}(a^{b}) = b, so we have the same thing here.0406
log_{a}(a) will drop us down to just having the b, so that b will move over to being the exponent, and we will have a^{b}.0410
But how did we define a^{b} in the first place? It was a^{b} = x.0418
So, we know that a^{b} = x; since we created b based on the fact that a^{b} = 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 log_{a}(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 log_{a}(x^{n}); by our key idea,0473
we know that there exists some b such that we can write a^{b} = xthe same thing.0478
So, if we want, we can swap out our x here for a^{b}.0484
So, we have a^{b} to the n now; from our rules about exponents, we know that a^{b} 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
log_{a}(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 a^{b} = x, so we could write that in its logarithmic form.0519
And we would show that log_{a}(x) = b, because a^{b} = 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 log_{a}(x) = b, which means that we can swap the b out here for the log_{a}(x).0539
And we have that log_{a}(x) here; we started with log_{a}(x^{n}),0548
and now we see that we can take this n here and move it out front, and we have n times log_{a}(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 log_{3}(3^{2}), by this property,0565
we see that we could take this 2 and move it out front, and we would have 2 times log_{3}(3).0574
Well, log_{3}(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, log_{3}(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"): log_{a}(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 log_{a}(a^{m}a^{n}).0636
Now, from our work about exponents, we know that a^{m} times a^{n} is the same thing as a^{m + n}.0642
The same base means that we can add the exponents, so we have log_{a}(a^{m + n}).0649
And then, since it is log_{a} on an exponent base of a, we get cancellation once again through inverses;0655
log_{a} 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 a^{m} = M; that was how we set this up.0668
So, log_{a}(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 a^{n} = N; we can express it, instead, as log_{a}(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, log_{a}(M) here becomes here, and log_{a}(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, log_{a}(MN), is going to be the same thing as log_{a}(M) + log_{a}(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 log_{10}; 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 log_{10}(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, log_{a}(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 log_{a}(M); and here is the N^{1}, so we have log_{a}(N^{1}).0855
So, log_{a}(M)...and then, we also have our rule that we can bring down exponents.0861
log(x^{n}) 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 log_{a}(M)  log_{a}(N).0873
Thus, log_{a}(M/N) is equal to log_{a}(M)  log_{a}(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 log_{2}(32/2), we could write that, by our new rules, as log_{2}(32)0899
(is the part on the top), and then the part on the bottom is 2, so minus log_{2}(2).0911
log_{2}(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 5so we raise 2 to the 5, and we get 32; so log_{2}(32) is 5.0924
What is log_{2}(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 first32/2?0948
Well, we could write that as log_{2}(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
Sorrythere 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 log_{a}(M + N) is equal to log_{a}(M) + log_{a}(N).1002
or that log_{a}(M  N) is equal to log_{a}(M)  log_{a}(N).1007
Those are not truenot 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 signyou 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, log_{a}(a^{x}) = 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 log_{a}(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(x^{n}), 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
log_{a}(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 logthe idea that log_{a}(x) = y means that a^{y} = 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, a^{y}, 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: a^{b} = x, for any x that we wanted to get to.1171
So, these two ideasyou 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 log_{7}(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 log_{3}(81) and log_{9}(81).1304
log_{3}(81): well, we could rewrite 81 as 3^{4}, so we see that that is just 4 over here.1308
Now, log_{9}(81) we could rewrite, also, as 9^{2}; so with a base 9, that would come out as 2.1315
So, we see that log_{3}(81) = 2log_{9}(81), because 4 is equal to 2 times 2.1323
So, we see that log_{3}(81) is equal to 2 times log_{9}(81).1333
And this is somehow related to the fact that 3^{2} 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 log_{3}(x) = 2log_{9}(x).1350
But let's prove it, instead of just assuminginstead 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 log_{9}(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 log_{9}(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 log_{3} of both sides.1396
We know that x is equal to 9^{log9(x)}, so it must be the case that log_{3}(x) is the same thing as 9^{log9(x)}.1398
So, log_{3}(9^{log9(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, log_{3}(x) is equal to log_{3}(9^{log9(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 log_{9}(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 log_{3}(x) is equal to log_{9}(x) times log_{3}(9).1437
Now, what is log_{3}(9)? That comes out to be 2, so we can simplify that as log_{3}(x) = 2log_{9}(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 = v^{logv(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 log_{u}, because we want to get v and u.1465
We want to get both of those logs into action, so we take log_{u} on both sides.1471
And then, we can bring this down in front, because it is an exponent; and we have log_{u}(x) = log_{v}(x) times log_{u}(v).1475
At this point, we can create a formula that will have log_{v} on one side and log_{u} on the other side.1483
We divide log_{u}(v) over, and we rearrange; we swap the order of the equation.1488
We have log_{v}(x) = log_{u}(x)/log_{u}(v).1493
Notice that this allows us to change from an expression log_{v}(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 havingif it is log_{7}(42), then we can change it into log_{10}(42)/log_{10}(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 log_{a}(M)  log_{a}(N).1558
The same log base is on both, and it splits into subtraction.1570
So, in this case, we have, on the top, x^{5}; so we will have log_{5}, the same thing,1573
minus what is on the bottom, y√z, so  log_{5}(y√z).1580
Great; now, we also have the rule that log_{a}, for any a of M times N, equals log_{a}(M) + log_{a}(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
log_{5}(x^{5}) ...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, log_{a}(M)  log_{a}(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 log_{5}(x^{5})  our subtraction distributes...minus here, as well... log_{5}(√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 log_{a},1668
for any base, of x^{n}, is equal to n times log_{a}(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 5log_{5}(x).1695
We continue to just bring down our log_{5}(y) ...we will rewrite the z...log_{5}(z^{1/2}).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 5log_{5}(x)  log_{5}(y)  1/2log_{5}(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(a^{3}) + 2(ln(b^{2}))  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(a^{1/3}) + 2(ln(b^{2}))  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 b^{2}/c; subtraction of logs is the same thing as division inside of the logarithm.1812
And now, natural log a^{1/3} 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(b^{2}/c)...make sure we remember that we are doing the whole logarithm of that whole thing,1834
so ln(a^{1/3}) + ln...we distribute that...(b^{4}/c^{2}).1844
And we can also bring this in now: natural log of...addition of logarithms becomes multiplication inside of the logarithms.1853
So, a^{1/3} times the rest of it...b^{4}...it will show up in the numerator, divided by c^{2}.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 log_{7}(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 log_{7}(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 log_{v}(x) = log_{u}(x)/log_{u}(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 log_{10}(42)/log_{10}(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 workwe want to make sure that that did work out1986
we would check it by saying that 7^{1.9208} 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 log_{π}(√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 log_{10}(√17)/log_{10}(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 withthey 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 log_{5}(a) = 6, and log_{5}(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 log_{5}(a) before we can swap it out for 6; so we have to get that sort of thing to show upthe same for log_{5}(b).2068
So, we split things up; we have multiplication between each one of thesethat is what it means when they are just stacked on top of each other.2075
So, log_{5}(5) + log_{5}(a) + log_{5}(b^{3}):2084
log_{5}(5) is just 1, because it is the same thing as its base, and what it is operating on,2095
plus log_{5}(a); we were told that was 6, so we get 6;2100
plus...now, log_{5}(b^{3})...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 3log_{5}(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: log_{25}(√a); this one is a kind of a problem.2139
We have 25 here, but we were told our base for working with the stuffthe 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 log_{25}(√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; log_{5}(√a)...now, √a is not what we havewe have a.2184
Is there another way to write √a that would involve a?2193
Yes, √a is the same thing as a^{1/2}: so we can write that as log_{5}(a^{1/2}).2196
divided by log_{5}(25)...well, we see that that is log_{5}(5^{2}),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, 1/2log^{5}(a)/2log_{5}(5)...log_{5}(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 gocool.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: log_{3}(x + 5) = 2.2260
Now, remember: we have that inverse propertywe know that a to the log_{a} 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 log_{3}(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 log_{3}(x + 5) over here.2301
3^{2} and 3^{log3(x + 5)}: 3 and log_{3} end up canceling out,2309
and x + 5 just drops down; that equals...there is nothing to cancel on the right side; it is 3^{2};2325
but we know what 3^{2} 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 log_{3}(4 + 5) = 2; does it?2342
log_{3}(4 + 5)let's see if that ends up coming out to be 2.2352
log_{3}(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; e^{x  8} = 47: to do this one, we remember that log_{a}(a^{x}) = 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 log_{e}, 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 e^{x  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(e^{x  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 notsimple 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 e^{3.85}; e^{3.85} 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 latergoodbye!2551
1 answer
Last reply by: Professor SelhorstJones
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 SelhorstJones
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 SelhorstJones
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 SelhorstJones
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 SelhorstJones
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