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### Course: Algebra 2 > Unit 8

Lesson 3: Properties of logarithms- Intro to logarithm properties (1 of 2)
- Intro to logarithm properties (2 of 2)
- Intro to logarithm properties
- Using the logarithmic product rule
- Using the logarithmic power rule
- Use the properties of logarithms
- Using the properties of logarithms: multiple steps
- Proof of the logarithm product rule
- Proof of the logarithm quotient and power rules
- Justifying the logarithm properties

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# Intro to logarithm properties (2 of 2)

Sal introduces the logarithm identities for multiplication of logarithm by a constant, and the change of base rule. Created by Sal Khan.

## Want to join the conversation?

- The exercise "Logarithms 2" is asking me to find log(3) + log(5).

This confuses me because there's no number above them, and because they have different bases. Can someone please explain to me why this is?(33 votes)- Hello Meredith. When "log" is written without subscripts (little numbers below the word log) it is assumed to be base 10. The 10 is left out. (Like the positive sign in positive numbers). So here you are adding two logarithms with base 10. As Sal explains log(a)+log(b) = log (a times b). The answer here is log(15).(81 votes)

- I hate memorizing. I love understanding. Please someone help me understand how did the exponent convert into a coefficient, and vice-versa. Thanks(17 votes)
- log_b (x^e) = y [Let]

So, x^e = b^y

So, (x^e)^(1/e) = (b^y)^(1/e) = b^(y/e)

So, x = b^(y/e)

So, log_b (x) = y/e

So, e log_b (x) = y = log_b (x^e)

This is the formal proof.

Taking a simpler example using a previously learnt property,

log_b (x^5)

= log_b (x.x.x.x.x) = log_b (x) + log_b (x) + log_b (x) + log_b (x) + log_b (x)

= 5 log_b (x)

and 5 is the original exponent of x. Hence proved.(19 votes)

- I lost him where he appears to have reduced 1/2log2,32 to 5/2, at9:51. Help.(9 votes)
- This is because the 1/2 cancels out the outer square root, but it doesn't affect the square root of 8 because the 8 is a square root inside the outer square root.(6 votes)

- At4:39, what does C stand for?

Those are awesome videos. Keep making more.(10 votes)- It is simply a variable for the possible number that can go there.(10 votes)

- also, do we always assume that "C" in the 2nd property of this video will be base 10 or can we just throw any number in for "C"?(6 votes)
- It can be any number however if you have a calculator it will have a log base 10 button built in which makes using 10 as C easier.(5 votes)

- how to solve: log base 3 of 9x^4 - log base 3 of (3x)^2(4 votes)
- Firstly, you cannot solve an expression, but you can simplify it like this:

log base 3 of 9x^4 - log base 3 of 9x^2 =

log base 3 of (9x^4/9x^2) =

log base 3 of (x^4/x^2) =

log base 3 of (x^2)(7 votes)

- Oof he kinda lost me with that last example

Great videos though! Really appreciate it(4 votes)- Assuming you are referring to the example

log_2(sqrt(32 / sqrt(8)))

There are different ways to do this, but I will follow the way the video described.

First, you need to know the following 2 formula / property

sqrt(a) = a^(1 / 2)

Simply definition of square root.

log(a^n) = n * log(a)

Mentioned at the beginning of the video.

Now we can rewrite our initial equation as

log_2((32 / sqrt(8))^(1 / 2))

= (1 / 2) * log_2(32 / sqrt(8))

Then, they make use of the following log property

log(a / b) = log(a) - log(b)

Mentioned in the last video Part 1.

We can continue our calculation.

(1 / 2) * log_2(32 / sqrt(8))

= (1 / 2) * (log_2(32) - log_2(sqrt(8)))

= (1 / 2) * (log_2(32) - log_2(8^(1 / 2)))

= (1 / 2) * (log_2(32) - (1 / 2) * log_2(8))

= (1 / 2) * (5 - (1 / 2) * 3)

= 1.75

If you still have any question, feel free to ask.(5 votes)

- i used base 17 instead of 10 in the calculator yet still got the right answer, why is that? then whats the use of the property at4:20?(4 votes)
- I'm not sure what you mean, because log base 10 of 357 is not the same as log base 17 of 357. The property is how you input other bases of logarithms into calculators though, so I'm not sure how you used base 17.(3 votes)

- I do not understand how you took the exponent "(1/2)" from log2(32/sqrt(8))^1/2, moved it to the left side of the equation, and then used it to multiply. I thought it is still an exponent, how did you use it to multiply the final result to get 5/2-3/4?(5 votes)
- Is there a formal proof video of these logarithm properties on Khan Academy?(3 votes)
- If you are on the algebra 2 playlist there should be proof videos near the bottom of it.

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:log-prop/v/proof-log-a-log-b-log-ab(4 votes)

## Video transcript

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!