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### Course: Integrated math 3 > Unit 5

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 product rule

Sal proves the logarithm addition property, log(a) + log(b) = log(ab). Created by Sal Khan.

## Want to join the conversation?

- Either I ams tupid, or Sal is just using the rule as its own proof.

It's like saying 2+2=4, and since 2+2=4, we can deduct 4=2+2, thus 2+2=4.

Am I missing something? Please help explain.(34 votes)- It's almost like you say, because it's so simple that the proof seems unnecessary:

Since you add exponents of powers of the same base to get the exponent of the product, and since logs are exponents, the log_b of the product is the sum of the logs_b of the factors.(8 votes)

- Though it has not been mentioned in the video , what are anti-logarithms?(9 votes)
- An anti-logarithm is essentially exponentiation. For example:

log (base 10) 5 = .6989700043

10^.6989700043 = 5(8 votes)

- Did Sal really refer to his colors 6 times in a single video?(10 votes)
- Why is the video so blurry?

Pls make another one(9 votes)- Because it was made it 2008.(3 votes)

- At around4:50, you say that x^l * x^m = x^(l+m)

I thought it would be 2x^(l+m)

WHat am I missing?(5 votes)- Where would you get the 2 from? You are multiplying, not adding.

Try this with some actual number, maybe that might make the concept more clear:

4² * 4³ = 4*4 * 4*4*4 = 4⁵ = 1024(7 votes)

- Feels like he did this one on an etch a sketch. Can't read any of that ;-x(8 votes)
- how is this concept useful guys somebody help?(4 votes)
- In science and engineering, there are a lot of phenomena that are exponential in nature. The variables are in the exponent, not in the base. In order to solve for the variables you have to take the log of the exponential functions. This is actually used a lot.(5 votes)

- Rather than worry about the colors, I wish Sal would focus on the legibility.(5 votes)
- Yeah, the audio is kind of stuffy too.(1 vote)

- I know this is wrong but I can't think out of it.

So if (LogX(A)=l) + (LogX(B)=m) = (LogX(A*B)=n) by the product rule. And if were to be converted to exponential form, would it look like this: (X^l=A) + (X^m=B) = (X^n=A*B), and if it is, then wouldn't this be equivalent to A+B=A*B which doesn't sound right.(3 votes)- Let's go through the correct application of the logarithmic properties and show why the statement is incorrect:

The product rule for logarithms states that log_x(A) + log_x(B) = log_x(A * B).

Suppose we have the expressions: (LogX(A) = l) and (LogX(B) = m).

According to the product rule, combining these two expressions should give us:

log_x(A) + log_x(B) = log_x(A * B).

However, we cannot directly add the two logarithmic expressions (log_x(A) and log_x(B)) as if they were numerical values (like "l" and "m").

To convert to exponential form, we would use the following:

log_x(A) = l -> x^l = A

log_x(B) = m -> x^m = B

Then, we can apply the product rule to the exponential forms:

x^l * x^m = A * B

Using the property x^a * x^b = x^(a+b):

x^(l+m) = A * B

However, we cannot say that (x^l = A) + (x^m = B) = (x^(l+m) = A * B).

Your observation that this would lead to A + B = A * B is indeed correct, but that's because the manipulation of logarithmic expressions in this way is not valid.

It's essential to use logarithmic properties correctly and to remember that logarithms do not follow the same arithmetic rules as regular numbers. When dealing with logarithms and their properties, it's crucial to apply them correctly to avoid incorrect conclusions or statements.

"Never back down, never give up"- Nick Eh30(3 votes)

- This is the proof assuming A and B are multiples of x. Is there a proof assuming they are not? The way power rule was proved is because it was dependent on product rule to be true. I was able to prove that product rule is true even if A and B are not multiples of x by using the power rule. But does this not cause contradiction? Proving a rule D then proving rule Z by proving it with D. However proving rule D by using rule Z sort of creates a paradox. Where the proof sort of collapses on itself, right? But not if I assume it is a special case of D where a and b are not multiples of a single base which I have proven, independently so I use the power rule which relies on a and b being multiples. Abstract math is way more fun than ordinary math. It makes math more clear. Thanks for all of the proofs Sal!(3 votes)

## Video transcript

Hello. Let's do some work on
logarithm properties. So, let's just review real
quick what a logarithm even is. So if I write, let's say I
write log base x of a is equal to, I don't know,
make up a letter, n. What does this mean? Well, this just means that
x to the n equals a. I think we already know that. We've learned that in
the logarithm video. And so it is very important to
realize that when you evaluate a logarithm expression, like
log base x of a, the answer when you evaluate, what
you get, is an exponent. This n is really
just an exponent. This is equal to this thing. You could've written
it just like this. You could have, because this n
is equal to this, you could just write x, it's going to get
a little messy, to the log base x of a, is equal to a. All I did is I, took this n and
I replaced it with this term. And I wanted to write it this
way because I want you to really get an intuitive
understanding of the notion that a logarithm, when
you evaluate it, it really an exponent. And we're going to
take that notion. And that's where, really,
all of the logarithm properties come from. So let me just do -- what I
actually want to do is, I want to to stumble upon the
logarithm properties by playing around. And then, later on, I'll
summarize it and then clean it all up. But I want to show maybe
how people originally discovered this stuff. So, let's say that x,
let me switch colors. I think that that keeps
things interesting. So let's say that x to
the l is equal to a. Well, if we write that as
a logarithm, that same relationship as a logarithm, we
could write that log base x of a is equal to l, right? I just rewrote what I
wrote on the top line. Now, let me switch colors. And if I were to say that x to
the m is equal to b, it's the same thing, I just
switched letters. But that just means that
log base x of b is equal to m, right? I just did the same thing
that I did in this line, I just switched letters. So let's just keep going
and see what happens. So let's say, let me
get another color. So let's say I have x to the n,
and you're saying, Sal, where are you going with this. But you'll see. It's pretty neat. x to the
n is equal to a times b. x to the n is equal
to a times b. And that's just like
saying that log base x is equal to a times b. So what can we do
with all of this? Well, let's start with
with this right here. x to the n is equal
to a times b. So, how could we rewrite this? Well, a is this. And b is this, right? So let's rewrite that. So we know that x to
the n is equal to a. a is this. x to the l. x to the l. And what's b? Times b. Well, b is x to the m, right? Not doing anything
fancy right now. But what's x to the
l times x to the m? Well, we know from the
exponents, when you multiply two expressions that have the
same base and different exponents, you just
add the exponents. So this is equal to, let
me take a neutral color. I don't know if I said that
verbally correct, but you get the point. When you have the same base and
you're multiplying, you can just add the exponents. That equals x to the, I want to
keep switching colors, because I think that's useful. l, l plus m. That's kind of onerous to
keep switching colors, but. You get what I'm saying. So, x to the n is equal
to x to the l plus m. Let me put the x here. Oh, I wanted that to be green. x to the l plus n. So what do we know? We know x to the n is equal
to x to the l plus m. Right? Well, we have the same base. These exponents must
equal each other. So we know that n is
equal to l l plus m. What does that do for us? I've kind of just been playing
around with logarithms. Am I getting anywhere? I think you'll see that I am. Well, what's another
way of writing n? So we said, x to the n is
equal to a times b -- oh, I actually skipped a step here. So that means -- so, going
back here, x to the n is equal to a times b. That means that log base x
of a times b is equal to n. You knew that. I didn't. I hope you don't realize I'm
not backtracking or anything. I just forgot to write that
down when I first did it. But, anyway. So, what's n? What's another way
of writing n? Well, another way of
writing n is right here. Log base x of a times b. So, now we know that if we just
substitute n for that, we get log base x of a times b. And what does that equal? Well, that equals l. Another way to write
l is right up here. It equals log base
x of a, plus m. And what's m? m is right here. So log base x of b. And there we have our
first logarithm property. The log base x of a times b --
well that just equals the log base x of a plus the
log base x of b. And this, hopefully,
proves that to you. And if you want the intuition
of why this works out it falls from the fact that logarithms
are nothing but exponents. So, with that, I'll leave
you with this video. And in the next video,
I will prove another logarithm property. I'll see you soon.