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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m

0. Functions

Properties of Logarithms

0. Functions

Properties of Logarithms - Online Tutor, Practice Problems & Exam Prep

1

concept

Evaluate Logarithms

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Hey, everyone. We just learned that a logarithm is the inverse of an exponential, but who cares, and how is that going to help us? Well, you're going to be asked to evaluate logarithms, like this log base 2 of the cube root of 2, without using a calculator. And looking at that, you might think it's impossible to get an answer there without using a calculator.

But here I'm going to show you that this log223 is actually just equal to 1 third. And we can get that answer simply by using the fact that the log is the inverse of an exponential. So with that in mind, let's go ahead and get started. Now we're going to look at a couple of different properties here, and the name of these properties is not important, just that we know how to use them. So let's get started with our first one, the inverse property.

Now here we see log223 and the first thing you might notice here is that the base of my log and the base of my exponent are the same. Now whenever that happens, these are simply going to cancel out, leaving me with just that 3. It's kind of similar to if I take the square root of something squared, I'm just left with that something. This is the same idea here. Now the same thing happens if I take an exponent of some base and raise it to a log of that same base.

They are also going to cancel, just leaving me with whatever is left over, in this case, 3. And this is because logs and exponentials of the same base are always going to cancel because they are each other's inverse. Now this will happen no matter what the base is. If I take logeex, I would still simply be left with x. Or if I took e and raised it to the power of loge, as long as those bases are the same, it doesn't matter.

They're going to cancel, leaving me with whatever is left over. So let's look at another property here. Here I have log22. Now here we see that the base of our log is the same thing that we're taking the log of. And we can think about this kind of similarly to our previous property where this 2 is actually the same thing as 2 to the power of 1 because 2 to the power of 1 is just 2.

Now with that in mind, using our inverse property, I know that this log of base 2 and my exponent of base 2 are going to cancel, leaving me with just 1. Now this is actually going to work for any log of any base. If I take the log of some base and I'm taking the log of that same number that is the base, I'm simply going to be left with 1 because taking the log of its base always equals 1 because of our inverse property. Now let's look at one final property here. Here I have log21.

Now here it's going to be helpful to think about this in its exponential form. So if I take my log of base 2 and think about what power I need to raise it to in order to equal 1, so 2x=1, I just want to think about what number I could plug in for x that would actually give me this one. So I know that 21=2, so it's not that. And I know that 22=4, so I actually need a lower number. But if I take it to 20=1, I know that anything0=1.

So here my answer would simply be 0. And this will be true any time we take the log of any base of 1. So any log of 1 is always going to be equal to 0 no matter what. Now with these properties in mind, let's take a look at some examples down here. Now looking at our first example, we already know what the answer is, but let's figure out how to get to that answer.

So log2213, I want to think about how I can rewrite this in a way that something is going to cancel. Now, this cube root of 2, I know that I can actually rewrite this as an exponent because the cube root of 2 is actually just 213 using our exponent rules to make that cube root into an exponent. So this is really log2213. Now with that in mind, using our inverse property, I know that this log base 2 and 2 cancel simply leaving me with 1 third, which is my answer, no calculator needed. Now let's move on to our next example here.

We have the natural log of 1. Now I know that the natural log is really log base e, so it's still a log. And I know that any time I take the log of any base of 1 I'm simply going to end up with 0. So that's my answer here, just 0. Let's look at another example here.

We have log of 10. Now log by itself is the common log so this is really log1010. And since the base is the same as what I'm taking the log of, using this second property up here, I know that this is simply going to be equal to 1 and I'm done here. Now we have one final example here. We have this log515.

And looking at this first glance, I'm not really sure exactly how I'm going to get an answer here. So we're going to have to be a little bit clever sometimes in thinking about how we can manipulate this in order to get something to cancel and get a final answer. Now since I have this log base 5, I know that I probably want my base of my exponent or what I'm taking the log of to be 5. So how can I rewrite this one fifth here? Well, I know that if I have a fraction, I could rewrite it as that number to the power of negative one.

So this is really log55-1. So we had to be a little bit clever there. But now that we're here, we can go ahead and just cancel some stuff out. So this log base 5 and 5 cancel, leaving me with just that negative one, which is my final answer, negative one. So now that we know how to evaluate logs with no calculator needed just using some inverse properties, let's get some more practice.

2

Problem

Problem

Evaluate the given logarithm.
$\log_77^{0.3}$

A

$1.79$

B

$7$

C

$1$

D

$0.3$

3

Problem

Problem

Evaluate the given logarithm.
$\frac32\log1$

A

$\frac32$

B

$0$

C

$1$

D

$10$

4

Problem

Problem

Evaluate the given logarithm.
$\log_9\frac{1}{81}$

A

$81$

B

$2$

C

$-2$

D

$9$

5

concept

Product, Quotient, and Power Rules of Logs

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3m

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Hey, everyone. When working through problems dealing with logs, you're going to come across questions that ask you to expand log expressions, like, say, log23x, into multiple different logs. And we can do that using certain properties of logs. Now don't worry. We're not just going to have to memorize a bunch of brand new rules here because all of these properties of logs actually correspond to properties of exponents that we already know and have used before.

So the same way we were able to graph log functions using what we knew about exponential functions, we can do the same thing here. So I'm going to give you a quick refresher of these exponential rules and then show you their corresponding log rule and how to use that to expand log expressions. So let's go ahead and jump right in. Now remember when working with rules like this, we don't really care about the name. That name is not important.

It's just important that we know how to use these rules, and the name is just a way to organize them. So starting first with our product rule here, remember that whenever we had exponents of the same base being multiplied together, we could simply take those exponents and add them together. And we see something similar when working with logs. So if I have some log of some base with 2 things being multiplied together, I can separate that out into 2 different logs that are simply being added together. So with our exponent, we saw multiplication turn to addition and we see the same thing happening here.

Whenever we have terms being multiplied in a log, this simply means that we can add 2 different logs together. Now we see something similar happen when working with the quotient rule because with our exponents, we saw that whenever we had exponents of the same base being divided, we would simply take those exponents and subtract them. So when seeing a log with 2 things being divided, what do you think I'm going to do to 2 separate logs? Well, I'm going to end up subtracting them. So the same way that we had division turn to subtraction with our exponent, the same thing is going to happen with our logs.

So whenever we divide terms in a log, we see that we subtract 2 logs. So multiplication becomes addition. Division becomes subtraction. So if I see something like log23x, like I have right here, I see this 3 and this x are being multiplied together. So since those things are being multiplied, I can turn this into the addition of 2 logs.

So this would become log base 2 of 3 plus log base 2 of x. So, see, you notice that that base stays the same, but I'm simply adding 2 logs together and I have separated that 3 and that x into that addition. Now if I'm given something like log55y, since that 5 and that y are being divided, this division turns to the subtraction of 2 separate logs. So this becomes log base 5 of 5 minus log base 5 of y using that quotient rule. Now we have one final log property to look at here, the power rule.

And whenever we worked with exponents, we saw that if we took an exponent of some base, so b to the power of m, and raised that to another power like n, I would simply end up multiplying those 2 powers together. Now we see something slightly different when working with our log, but it's still going to see multiplication happening. So if I have logbmn, it's still going to turn out into multiplication. But I'm going to take this n and stick it on the front of that log. So log base m to the power of n becomes n times log base b of m.

So I simply take that n and pull it out to the front and multiply it together. So anytime I'm raising a term to a power I am simply going to end up multiplying the log by the power. So if I'm given something like the natural log of 7 to the power of 2, I'm going to take that 2 and stick it on the front of that. And this is going to become 2 times the natural log of 7.

So now that we've seen these log properties, we have the tools we need to expand log expressions. Thanks for watching, and I'll see you in the next one.

6

concept

Expand & Condense Log Expressions

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4m

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Hey, everyone. We just learned all of the rules needed to expand log expressions, but two things are going to happen. We're going to be asked to expand much more complicated log expressions, like, say, log2(3xy2). And we're also going to be asked to do the exact opposite of expanding log expressions and actually condense expressions with multiple logs down to a single log. Now, you don't have to worry because we're going to take these rules that we already know, and I'm going to walk you through how to use multiple of these rules and how to use them in their reverse direction because each of these properties can be applied in both directions depending on what your goal is, whether taking a single log to multiple logs or condensing multiple logs back down into a single log.

Now you'll be able to expand and condense any log expression that gets thrown your way, so let's go ahead and get started. Now we're going to just jump right into an example here. And in our first example, we see that we have log2(3xy2), and we want to go ahead and expand this log expression as much as possible. So the first thing I notice here is that I have 3 different things being multiplied together. This is actually 3 times x times y2.

Now, since this is multiplication, that clues me in that I need to go ahead and use the product rule because I know that the product rule takes multiplication and turns it into the addition of multiple logs. So let's go ahead and expand this log out into multiple logs using that product rule. So this is log2(3) plus log2(x) plus log2(y2). Now I have these multiple logs, but I want to go ahead and walk through each of these single logs to make sure that I can't expand them any further. So I have log2(3).

I can't expand that anymore, so that's going to remain as a log2(3). And then log2(x), I also cannot expand that anymore, so that will remain the same as well. And then finally, we come to our last term, log2(y2). Now, looking at this log, I know that I have this exponent here. And whenever I have an exponent, that tells me that I can go ahead and use the power rule because the power rule tells me that I can take that exponent and pull it to the front of my log to expand it.

So this 2, I can go ahead and pull to the front of my log, and this last term becomes 2 times log2(y). Now none of these terms can be expanded anymore, so this is my final answer, log2(3) plus log2(x) plus 2 times log2(y). We've taken this one log and expanded it out as much as we can. Now let's look at condensing a log expression. Now, we do have to consider a couple of additional things whenever we're condensing logs.

And the first is that we always want to make sure that the base has to be the same. So here we have 2 times the natural log of x minus the natural log of x plus 2. So both of these are natural logs. They have that same base of e. If this was, say, the natural log and log base 2, I couldn't condense those logs together because they have different bases.

We'll always apply the power rule first to get the correct answer. So we have 2 times the natural log of x minus the natural log of x plus 2. I'm going to look at these terms and see how I can apply that power rule. So I have 2 times the natural log of x. I can take that thing that's multiplying it and make it into the exponent. So this becomes the natural log of x squared.

That second term, I can't apply the power rule, so it's going to remain the same for now, the natural log of x plus 2. How else can we condense this expression? They're being subtracted, so I should go ahead and use the quotient rule. I can take subtraction and turn it into the division of a single log expression. So this becomes the ln(x2×(x+2)). Now I have taken these multiple logs and condensed them down into a single log, so this is my final answer, the natural log of x2x+2. Now that we have seen how to expand and condense log expressions, let's go ahead and get some more practice. Thanks for watching, and let me know if you have questions.

7

Problem

Problem

Write the log expression as a single log.
$\log_2\frac{1}{9x}+2\log_23x$

A

$\log_2x$

B

$\log_2\frac{1}{3x}$

C

$\log_21$

D

$\log_23x$

8

Problem

Problem

Write the log expression as a single log.
$\ln\frac{3x}{y}+2\ln2y-\ln4x$

A

$\ln\frac{3xy}{4}$

B

$\ln\left(12x^2\right)$

C

$\ln\left(\frac32\right)$

D

$\ln\left(3y\right)$

9

Problem

Problem

Write the single logarithm as a sum or difference of logs.
$\log_3\left(\frac{\sqrt{x}}{9y^2}\right)$

A

$2 l o g_{3} x - 2 - l o g_{3} 9 y$

B

$\frac{1}{2} \log_{3} x - 2 - 2 \log_{3} y$

C

$one half l o g sub 3 of x plus 2 l o g sub 3 of 3 y$

D

$one half l o g sub 3 of x minus 2 l o g sub 3 of 9 y$

10

Problem

Problem

Write the single logarithm as a sum or difference of logs.
$\log_5\left(\frac{5\left(2x+3\right)^2}{x^3}\right)$

A

$5+2\log_5\left(2x+3\right)-\log_53x$

B

$2\log_5\left(2x+3\right)-3\log_5x$

C

$1+2\log_5\left(2x+3\right)-3\log_5x$

D

$\log_5\left(2x+3\right)-\log_5x$

11

concept

Change of Base Property

Video duration:

3m

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At this point, we've evaluated logs by hand and we've used different rules and properties in order to expand and condense log expressions. But sometimes you're just going to need to quickly evaluate a log by plugging it into your calculator. But if you're given something like log base pi of 9, well, that doesn't look quite so simple to just type in our calculator and go. So here I'm going to show you that whenever you're given a log with some base that doesn't look so easy to deal with, like, say, pi, you can simply change the base to be whatever you need it to be. So here I'm going to walk you through exactly how to do that and let's not waste any time here and just jump right in.

So if I have some log, log base b of m, but my base b is, say, pi or 27 or 500, some number that is not so easy to deal with, so I want to change it, I can go ahead and change it by taking my log and turning it into a fraction. So if my original base was b but I wanted a base of a, I would simply take log base a of m and divide it by log base a of b. So whatever I was originally taking the log of m is going to go on the top and then my original base b goes on the bottom. That's one way to remember this. Base goes on the bottom.

So let's say that I'm given some log, like log base 5 of 2, and I want to change that to have a base of 10. I could go ahead and change that into a fraction. And that fraction would be log base 10 of 2, whatever I was originally taking the log of, divided by log base 10 of 5. Now from here, since there actually is a button for log base 10, my common log on my calculator, I can quickly and easily type this in my calculator and come up with an answer. Now because log base 10, your common log, has a button on your calculator and so does the natural log, log base e, you're most often going to want to change your base to be either 10 or e so that you can quickly type it in your calculator and get an answer.

So we just saw that whenever we changed our base to 10, this would become log base 10 of m divided by log base 10 of b. Now, log base 10 can, of course, just be written as log because it's the common log. So this is really just log of M divided by the log of b. Now we see the same thing whenever we're faced with changing our base to e. We know that log base e is just the natural log.

So this could simply be written as the natural log of M divided by the natural log of b. Now you can change any log of any base to either of these two options. Most often, it will be specified to you which one you want to change it into. But if it's not, you can use either one, and it doesn't really matter. So with that in mind, let's go ahead and work through some examples here.

So with our first example, we see log base 7 of 31. So log base 7 of 31 and for a and b here. We want to go ahead and use common logs, so changing our base to 10. So looking at my first base here, log base 7 of 31, I want to change this to log base 10. So doing that, I can take log base 10 in my fraction and just plug in whatever I need to plug in on the top and bottom.

Bottom. So I'm going to take whatever I was originally taking the log of on top. So this is log base 10 of 31 and then divided by log base 10 of my original base 7. Now, of course, log base 10 can just be written as log. So this is just log of 31 divided by the log of 7.

Now I can go ahead and type this into my calculator. And when I do, I'm going to go ahead and get an answer of 1.76, and I have fully evaluated that log. Now let's look at another example. Here we have log base pi of 9, and we want to again change this into a log base 10 because we're still using common logs here. So log base pi of 9, I know that I can turn this into log of 9 divided by log of pi.

So whatever I was originally taking the log of, goes on the top. And then, of course, my base goes on the bottom. So I can go ahead and plug this into my calculator, log of 9 over the log of pi. And I'll end up getting an answer of 1.92. And I've evaluated that log.

Now looking at our 3rd example, example c, we again have log base pi of 9. But now we want to go ahead and use a natural log, so changing our base instead to e. So knowing that log base e is just the natural log, I know that this will turn into the natural log of 9 divided by the natural log of pi. So again, whatever we were originally taking the log of goes on the top, and then my original base goes on the bottom. So I can go ahead and type this in my calculator, natural log of 9 divided by the natural log of pi, and I'm going to end up getting 1.92.

Now earlier I said it doesn't matter what you change your base into and that's because we're going to get the same answer regardless. Whenever we changed our base to 10, we got 1.92. And then whenever we changed our base to e, we also got 1.92. So it doesn't really matter. You're going to get the same answer regardless.

Let's take a look at one final example here. We have log base the square root of 3 of e. And I want to use natural logs for this one as well. So changing this to a base of e, I'm going to get the natural log of e divided by the natural log of the square root of 3. Now we can actually do some further simplification here.

Because the natural log we know is just log base e, simplification here because the natural log we know is just log base e, so this is really log base e of e. Now whenever we have the same base as what we're taking the log of, I know that this just simplifies to 1. So this really just becomes 1 over the natural log of the square root of 3. Now I can go ahead and just plug that into my calculator and when I do I'm going to get an answer of 1.82, and I have fully evaluated that log. Now that we know how to change the base of a log to be whatever we need it to be, let's get some more practice.

Let me know if you have any questions.

12

Problem

Problem

Evaluate the given logarithm using the change of base formula and a calculator. Use the common log.
$\log_317$

A

$0.39$

B

$2.58$

C

$1.23$

D

$0.48$

13

Problem

Problem

Evaluate the given logarithm using the change of base formula and a calculator. Use the common log.
$\log_967$

A

$1.91$

B

$0.52$

C

$0.95$

D

$1.83$

14

Problem

Problem

Evaluate the given logarithm using the change of base formula and a calculator. Use the natural log.
$\log_841$

A

$1.61$

B

$0.9$

C

$0.56$

D

$1.79$

15

Problem

Problem

Evaluate the given logarithm using the change of base formula and a calculator. Use the natural log.
$\log_23789$

A

$0.08$

B

$11.89$

C

$3.58$

D

$0.30$

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