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https://play.kahoot.it/#/k/8430381f-89fb-4495-a338-9ea8bc12559e

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Example 1
For this problem we had to use u-sub to find the anti-derivative. But once we found that, we could replace the u and use the given limits and FTC to solve the integral. However, at the end, you must know how to combine logs in order to get the simplest and final answer.

Example 2
We have to use u-sub again in this problem in order to find the anti-derivative of the given integral. After finding it, we are able to replace the u and solve for the answer with the given limits and FTC. We must use logarithm properties in order to combine the logs and get a final answer.

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https://www.slideshare.net/jaflint718/ap-calculus-bc-integration-multiple-choice-practice-solutions

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Example 1
₁∫2 2x³/(x⁴+1) dx            u=x⁴+1
du=4x³ dx
(1/2) du= 2x³ dx
=(1/2) ₁∫² du/u

= (1/2)ln |u| ]₁ 2
= (1/2)ln |x⁴+1| ]₁ 2
=(1/2)ln |2⁴+1| - (1/2)ln |1⁴+1|
= (1/2)ln |17| - (1/2)ln |2|
=(1/2)ln |17/2|

Example 2
₁∫2 dx/(20-x)                   u=20-x
du=-dx
= - ₁∫2 du/u
= -ln |u| ]₁ 2
= -ln |20-x| ]₁ 2
= -ln |20-2| - (-ln|20-1|)
= -ln |18| + ln|19|
= ln|19/18|

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Example 5 Solve
4e(1+3x)-9e(5-2x)=0
4e(1+3x)=9e(5-2x)
(e(1+3x))/(e(5-2x))=9/4
e(1+3x-(5-2x))=9/4
e(5x-4)=9/4
ln(e(5x-4))=ln(9/4)
5x-4=ln(9/4)
5x=4+ln(9/4)
x=1/5(4+ln(9/4))

Example 6 Solve
3+2ln(x/7+3)=-4
2ln(x/7+3)=-7
ln(x/7+3)=-7/2
e(ln(x/7+3))=e(-7/2)
x/7+3=e(-7/2)
x/7=-3+e(-7/2)
x=7(-3+e(-7/2))

Example 7 Solve
2ln(x(1/2))-ln(1-x)=2
ln((x(1/2))²)-ln(1-x)=2
ln(x)-ln(1-x)=2
ln(x/(1-x)=2
x/(1-x)=e²
x=e²-(e²)x
x+(e²)x=e²
x(1+e²)=e²
x=e²/(1+e²)

Example 8 Solve
log(x)+log(x-3)=1
log(x(x-3))=1
log(x²-3x)=1
x²-3x=10
x²-3x-10=0
(x-5)(x+2)=0
x=5
x cannot equal -2 because you cannot take the log of a negative number

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http://tutorial.math.lamar.edu/Classes/CalcI/ExpLogEqns.aspx
Example 1  Solve
7+15e(1-3x)=10
15e(1-3x)=3
e(1-3x)=1/5
ln(e(1-3x))=ln(1/5)
1-3x=ln(1/5)
-3x=ln(1/5)-1
x=-1/3(ln(1/5)-1)

Example 2 Solve
log(10(t^2-t))=log(100)
t²-t=2
t²-t-2=0
(t-2)(t+1)=0
t=2 t=-1

Example 3 Solve
x-xe(5x+2)=0
x(1-e(5x+2))=0
x=0
1-e(5x+2)=0
e(5x+2)=1
ln(e(5x+2))=ln(1)
5x+2=0
x=-2/5

Example 4 Solve
5(x²-4)=(x²-4)e(7-x)
5(x²-4)-(x²-4)e(7-x)=0
(x²-4)(5-e(7-x))=0
x²-4=0
(x+2)(x-2)=0
x=-2 x=2
5-e(7-x)=0
e(7-x)=5
ln(e(7-x))=ln(5)
7-x=ln(5)
x=7-ln(5)

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Khan Academy

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PROFESSOR: Welcome back. I'm going to show you the last two logarithm properties now. So this one-- and I always found this one to be in some ways the most obvious one. But don't feel bad if it's not obvious. Maybe it'll take a little bit of introspection. And I encourage you to really experiment with all these logarithm properties, because that's the only way that you'll really learn them. And the point of math isn't just to pass the next exam, or to get an A on the next exam. The point of math is to understand math so you can actually apply it in life later on and not have to relearn everything every time. So the next logarithm property is, if I have A times the logarithm base B of C, if I have A times this whole thing, that that equals logarithm base B of C to the A power. Fascinating. So let's see if this works out. So let's say if I have 3 times logarithm base 2 of 8. So this property tells us that this is going to be the same thing as logarithm base 2 of 8 to the third power. And that's the same thing. Well, that's the same thing as-- we could figure it out. So let's see what this is. 3 times log base-- what's log base 2 of 8? The reason why I kind of hesitated a second ago is because every time I want to figure something out, I implicitly want to use log and exponential rules to do it. So I'm trying to avoid that. Anyway, going back. What is this? 2 to the what power is 8? 2 to the third power is 8, right? So that's 3. We have this 3 here, so 3 times 3. So this thing right here should equal 9. If this equals 9, then we know that this property works at least for this example. You don't know if it works for all examples, and for that maybe you'd want to look at the proof we have in the other videos. But that's kind of a more advanced topic. But the important thing first is just to understand how to use it. Let's see, what is 2 to the ninth power? Well it's going to be some large number. Actually, I know what it is-- it's 256. Because in the last video we figured out that 2 to the eighth was equal to 256. So 2 to the ninth should be 512. So 2 to the ninth should be 512. So if 8 to the third is also 512 then we are correct, right? Because log base 2 of 512 is going to be equal to 9. What's 8 to the third? It's 64-- right. 8 squared is 64, so 8 cubed-- let's see. 4 times 2 is 3. 6 times 8-- looks like it's 512. Correct. And there's other ways you could have done it. Because you could have said 8 to the third is the same thing as 2 to the ninth. How do we know that? Well, 8 to the third is equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth. And actually it's this exponent property, where you can multiply-- when you take something to exponent and then take that to an exponent, and you can essentially just multiply the exponents-- that's the exponent property that actually leads to this logarithm property. But I'm not going to dwell on that too much in this presentation. There's a whole video on kind of proving it a little bit more formally. The next logarithm property I'm going to show you-- and then I'll review everything and maybe do some examples. This is probably the single most useful logarithm property if you are a calculator addict. And I'll show you why. So let's say I have log base B of A is equal to log base C of A divided by log base C of B. Now why is this a useful property if you are calculator addict? Well, let's say you go class, and there's a quiz. The teacher says, you can use your calculator, and using your calculator I want you to figure out the log base 17 of 357. And you will scramble and look for the log base 17 button on your calculator, and not find it. Because there is no log base 17 button on your calculator. You'll probably either have a log button or you'll have an ln button. And just so you know, the log button on your calculator is probably base 10. And your ln button on your calculator is going to be base e. For those you who aren't familiar with e, don't worry about it, but it's 2.71 something something. It's a number. It's nothing-- it's an amazing number, but we'll talk more about that in a future presentation. But so there's only two bases you have on your calculator. So if you want to figure out another base logarithm, you use this property. So if you're given this on an exam, you can very confidently say, oh, well that is just the same thing as-- you'd have to switch to your yellow color in order to act with confidence-- log base-- we could do either e or 10. We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. I still use this in my job. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know if I'll move vertically or horizontally. I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right? Because this is in a quotient. Minus the logarithm base 2 of the square root of 8. Right? Let's see. Well here once again we have a square root here, so we could say this is equal to 1/2 times log base 2 of 32. Minus this 8 to the 1/2, which is the same thing is 1/2 log base 2 of 8. We learned that property in the beginning of this presentation. And then if we want, we can distribute this original 1/2. This equals 1/2 log base 2 of 32 minus 1/4-- because we have to distribute that 1/2-- minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4 is equal to 7/4. I probably made some arithmetic errors, but you get the point. See you soon!

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PROFESSOR: Welcome back. I'm going to show you the last two logarithm properties now. So this one-- and I always found this one to be in some ways the most obvious one. But don't feel bad if it's not obvious. Maybe it'll take a little bit of introspection.

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Khan Academy

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Intro to logarithm properties (1 of 2) (video) | Khan Academy

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It is very easy to make mistakes and not find an answer while integrating if you do not know how to properly use log properties. Therefore, we would like to review log properties before learning integration.