For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

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### 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 loglog _{v}x =log _{u}xlog_{u}v

. _{v}x to an expression that only uses log_{u}. By using u = e or u=10, we can evaluate with any calculator.

### 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

### Math Analysis Online

### 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 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 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 x ^{a} times x^{b} = x^{a + b}--that we could add exponents--*0046

*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 something*0057

*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: 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 to--what 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 2--10 ^{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 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 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 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 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} = x--the 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 5--so 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 first--32/2?*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

*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 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 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, 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 log--the 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 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 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 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 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 having--if 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 expression*1739

*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 work--we want to make sure that that did work out--*1986

*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 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 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 up--the same for log_{5}(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, 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 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 log _{25}(√a) is the same thing*2162

*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 have--we 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 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: log _{3}(x + 5) = 2.*2260

*Now, remember: we have that inverse property--we 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 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 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 later--goodbye!*2551

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