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## Algebra 2

### 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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# Proof of the logarithm quotient and power rules

Sal proves the logarithm quotient rule, log(a) - log(b) = log(a/b), and the power rule, k⋅log(a) = log(aᵏ). Created by Sal Khan.

## Want to join the conversation?

- what grade is logs for?(7 votes)
- Logarithms are part of the High School Algebra Core Curriculum Standards(20 votes)

- I think I just gained several brain cells at3:08(11 votes)
- who can prove this to me?? how it can be answer of 1?

1/(log_a AB) + 1/(log_b AB) = 1(3 votes)- That's easy (but changing b to x since there is a subscript x character):

1/logₐ(ax) + 1/logₓ(ax)

= [ log(a) / log(ax)] + [ log(x) / log(ax) ]

= [ log(a) + log (x) ] / log(ax)

= log (ax) / log (ax)

= 1

Provided that both a and x are positive. It is undefined if either a or x is ≤ 0(14 votes)

- What's the point of proofs?(3 votes)
- The point is to prove that this rules are not made up and that they are true. Just as you could put in the numbers into a formula you could prove it by using the properties o a function/operator to find out how the numbers move.(7 votes)

- By knowing the previous two properties (product and power), you could prove the quotient property this way:

log(a) - log(b) = log(a) + (-1)log(b) = log(a) + log(b^-1) = log(a) + log(1/b) = log(a * 1/b) = log(a/b)(6 votes) - Where is the next video Sal mentions at7:41?(3 votes)
- There are more logarithm properties than this, they should be added to this section.(4 votes)
- Indeed there are way more rules, at least the ones I studied at school(2 votes)

- If a^b=c, then

(a^b)^x=c^x

then log_a(c^x)=bx

but b=log_a(c)

so log_a(c^x)=x(log_a(c))

right?(4 votes)- Yes pretty much. Your second line easily follows from the first. Then on the third line log_a(c) is b due to what was stipulated on the first line.

So in that case raising the (c) in log_a(c) to the x is equivalent to multiplying b by x, so log_a(c^x) is indeed bx.

Then on the fourth line b is log_a(c), again by virtue of what was stipulated on the first line.

It seems pretty straight to me. If it was going to go a little off base anywhere it probably would've been the third line. Maybe you could've had the fourth line as your third line, since for the third to work you need to build on the very statement that is only made explicit on the fourth.(1 vote)

- how would you solve: 12^log12(4)(2 votes)
- since 12 is being brought to a log power with a base the same as it, they cancel out. so 12^log12(4)=4 Keep in mind this only works because there are the same number in those two specific places.

To see that it works first set it up like an equation.

12^log12(4) = x

Now turn it into another log. so if it were 12^y = x you sould make it log12(x)=y. so here 12^log12(4)=x becomes log12(x) = log12(4).

x has to be 4 because if log12(x) = log12(4) log base 12 only gets to the number log12(4) equals when the number inside is 4, no other number can get the sam result. so x must be 4(4 votes)

- Can anyone prove that logaX=-logaX using log law 3?(2 votes)
- If logₐ(x)=-logₐ(x), then logₐ(x)=0. That has nothing to do with logarithms, zero is the only number that is its own negative.(2 votes)

## Video transcript

Let's see if we can stumble
our way to another logarithm property. So let's say that the log
base x of A is equal to B. That's the same thing as saying
that x to the B is equal to A. Fair enough. So what I want to
do is experiment. What happens if I multiply this
expression by another variable? Let's call it C. So I'm going to multiply both
sides of this equation times C. And I'll just switch colors
just to keep things interesting. That's not an x that's a C. I should probably just
do a dot instead. Times C. So I'm going to multiply both
sides of this equation times C. So I get C times log base x of
A is equal to-- multiply both sides of the equation--
is equal to B times C. Fair enough. I think you realize I
have not done anything profound just yet. But let's go back. We said that this is the
same thing as this. So let's experiment
with something. Let's raise this side
to the power of C. So I'm going to raise this
side to the power of C. That's a kind of caret. And when you type
exponents that's what you would use, a caret. So I'm going to raise
it to the power of C. So then, this side is x
to the B to the C power, is equal to A to the C. All I did is I raised both
sides of this equation to the Cth power. And what do we know about when
you raise something to an exponent and you raise that
whole thing to another exponent, what happens
to the exponents? Well, that's just exponent
rule and you just multiply those two exponents. This just implies that x to the
BC is equal to A to the C. What can we do now? Well, I don't know. Let's take the logarithm
of both sides. Or let's just write this--
let's not take the logarithm of both sides. Let's write this as a
logarithm expression. We know that x to the BC
is equal to A to the C. Well, that's the exact same
thing as saying that the logarithm base x of A to
the C is equal to BC. Correct? Because all I did is I rewrote
this as a logarithm expression. And I think now you
realized that something interesting has happened. That BC, well, of course, it's
the same thing as this BC. So this expression must be
equal to this expression. And I think we have another
logarithm property. That if I have some kind of a
coefficient in front of the logarithm where I'm multiplying
the logarithm, so if I have C log-- Clog base x of A,
but that's C times the logarithm base x of A. That equals the log
base x of A to the C. So you could take this
coefficient and instead make it an exponent on the term
inside the logarithm. That is another
logarithm property. So let's review what we know
so far about logarithms. We know that if I write-- let
me say-- well, let me just with the letters I've been using. C times logarithm base x of
A is equal to logarithm base x of A to the C. We know that. And we know-- we just learned
that logarithm base x of A plus logarithm base x of B is equal
to the logarithm base x of A times B. Now let me ask you a question. What happens if instead of
a positive sign here we put a negative sign? Well, you could probably figure
it out yourself but we could do that same exact proof that
we did in the beginning. But in this time we will
set it up with a negative. Let's just say that log
base x of A is equal to l. Let's say that log base
x of B is equal to m. Let's say that log base x of A
divided by B is equal to n. How can we write all of these
expressions as exponents? Well, this just says that
x to the l is equal to A. Let me switch colors. That keeps it interesting. This is just saying that x
to the m is equal to B. And this is just saying that
x to the n is equal to A/B. So what can we do here? Well what's another
way of writing A/B? Well, that's just the same
thing as writing x to the l because that's A,
over x to the m. That's B. And this we know from our
exponent rules-- this could also be written as x to the
l, x to the negative m. Or that also equals
x to the l minus m. So what do we know? We know that x to the n is
equal to x to the l minus m. Those equal each other. I just made a big
equal chain here. So we know that n is
equal to l minus m. Well, what does that do for us? Well, what's another
way of writing n? I'm going to do it up here
because I think we have stumbled upon another
logarithm rule. What's another way
of writing n? Well, I did it right here. This is another
way of writing n. So logarithm base x of
A/B-- this is an x over here-- is equal to l. l is this right here. Log base x of A is equal to l. The log base x of A minus m. I wrote m right here. That's log base x of B. There you go. I probably didn't
have to prove it. You could've probably tried
it out with dividing it, but whatever. But you know are hopefully
satisfied that we have this new logarithm property right there. Now I have one more logarithm
property to show you, but I don't think I have time to
show it in this video. So I will do it in
the next video. I'll see you soon.