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.

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.

Though it has not been mentioned in the video , what are anti-logarithms?

An anti-logarithm is essentially exponentiation. For example: log (base 10) 5 = .6989700043 10^.6989700043 = 5

At around 4:50, you say that x^l * x^m = x^(l+m) I thought it would be 2x^(l+m) WHat am I missing?

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

how is this concept useful guys somebody help?

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.

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.

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

Why we use _log_ and not any other notation for this? And Why can't we only use exponents? Yeah Logarithm is an interesting topic but why we want logs when we have exponents?

Using "log" is just convention. Try solving for 𝑥 without logs: 5ˣ = 3ˣ ⁺ ¹

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Algebra 2

Course: Algebra 2 > Unit 8

Lesson 3: Properties of logarithms

Proof of the logarithm product rule

Name: Proof of the logarithm product rule
Uploaded: 2011-02-20T16:27:09Z
Description: Sal proves the logarithm addition property, log(a) + log(b) = log(ab).

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

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yangzeyuliu
Posted 9 years ago. Direct link to yangzeyuliu's post “Either I ams tupid, or Sa...”
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.
Button navigates to signup pageComment on yangzeyuliu's post “Either I ams tupid, or Sa...”
(34 votes)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “It's almost like you say,...”
  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.
  Button navigates to signup page
  (4 votes)
Khubaib Abdullah
Posted 10 years ago. Direct link to Khubaib Abdullah's post “Though it has not been me...”
Though it has not been mentioned in the video , what are anti-logarithms?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Frost
  Posted 10 years ago. Direct link to Frost's post “An anti-logarithm is esse...”
  An anti-logarithm is essentially exponentiation. For example:
  
  log (base 10) 5 = .6989700043
  10^.6989700043 = 5
  Comment on Frost's post “An anti-logarithm is esse...”
  (6 votes)
Schweddy Balls
Posted 10 years ago. Direct link to Schweddy Balls's post “At around 4:50, you say t...”
At around
4:50
, you say that x^l * x^m = x^(l+m)
I thought it would be 2x^(l+m)

WHat am I missing?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Just Keith
  Posted 10 years ago. Direct link to Just Keith's post “Where would you get the 2...”
  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
  Comment on Just Keith's post “Where would you get the 2...”
  (6 votes)
Jesse Bevil
Posted 6 years ago. Direct link to Jesse Bevil's post “Feels like he did this on...”
Feels like he did this one on an etch a sketch. Can't read any of that ;-x
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(6 votes)
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Chad willson
Posted 6 years ago. Direct link to Chad willson's post “Did Sal really refer to h...”
Did Sal really refer to his colors 6 times in a single video?
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(5 votes)
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o.mcqueen
Posted 12 years ago. Direct link to o.mcqueen's post “how is this concept usefu...”
how is this concept useful guys somebody help?
Button navigates to signup pageComment on o.mcqueen's post “how is this concept usefu...”
(3 votes)
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- Vince Mashburn
  Posted 8 years ago. Direct link to Vince Mashburn's post “In science and engineerin...”
  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.
  Button navigates to signup page
  (4 votes)
claudiapankaj
Posted 9 months ago. Direct link to claudiapankaj's post “Why is the video so blurr...”
Why is the video so blurry?
Pls make another one
Button navigates to signup pageComment on claudiapankaj's post “Why is the video so blurr...”
(4 votes)
Answer
unspecifiedunit
Posted 3 months ago. Direct link to unspecifiedunit's post “I know this is wrong but ...”
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.
Button navigates to signup pageComment on unspecifiedunit's post “I know this is wrong but ...”
(2 votes)
Answer
- him
  Posted 2 months ago. Direct link to him's post “Let's go through the corr...”
  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
  Button navigates to signup page
  (2 votes)
Yuvraj Singh Jadon
Posted 7 years ago. Direct link to Yuvraj Singh Jadon's post “Why we use _log_ and no...”
Why we use log and not any other notation for this?
And
Why can't we only use exponents?
Yeah Logarithm is an interesting topic but why we want logs when we have exponents?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- andrewp18
  Posted 7 years ago. Direct link to andrewp18's post “Using "log" is just conve...”
  Using "log" is just convention.
  Try solving for 𝑥 without logs:
  5ˣ = 3ˣ ⁺ ¹
  Button navigates to signup page
  (2 votes)
Taylor Bosick
Posted 11 years ago. Direct link to Taylor Bosick's post “if you had the equation 4...”
if you had the equation 4^x-2=15^-2x , how could you solve for x so that you would get the exact answer using base 10 logarithms? please help me using this example I cannot wrap my head around it! Thank you!
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(2 votes)
Answer

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.