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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.

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  • aqualine ultimate style avatar for user yangzeyuliu
    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.
    (35 votes)
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  • spunky sam blue style avatar for user Khubaib Abdullah
    Though it has not been mentioned in the video , what are anti-logarithms?
    (8 votes)
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  • duskpin ultimate style avatar for user claudiapankaj
    Why is the video so blurry?
    Pls make another one
    (10 votes)
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  • leaf green style avatar for user Schweddy Balls
    At around , you say that x^l * x^m = x^(l+m)
    I thought it would be 2x^(l+m)

    WHat am I missing?
    (5 votes)
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  • old spice man green style avatar for user Chad willson
    Did Sal really refer to his colors 6 times in a single video?
    (8 votes)
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  • male robot johnny style avatar for user Jesse Bevil
    Feels like he did this one on an etch a sketch. Can't read any of that ;-x
    (7 votes)
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  • mr pants teal style avatar for user o.mcqueen
    how is this concept useful guys somebody help?
    (3 votes)
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  • blobby green style avatar for user unspecifiedunit
    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)
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    • female robot amelia style avatar for user him
      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)
  • blobby green style avatar for user burnschurchill
    Rather than worry about the colors, I wish Sal would focus on the legibility.
    (4 votes)
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  • blobby green style avatar for user bakula.darko4
    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)
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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.