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## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 3

Lesson 5: Topic B: Lessons 10-12: Logarithm properties

# 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?
• Logarithms are part of the High School Algebra Core Curriculum Standards
• who can prove this to me?? how it can be answer of 1?
1/(log_a AB) + 1/(log_b AB) = 1
• 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
• I think I just gained several brain cells at
• 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)
• What's the point of proofs?
• 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.
• There are more logarithm properties than this, they should be added to this section.
• Indeed there are way more rules, at least the ones I studied at school
• 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?
• 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)
• Where is the next video Sal mentions at ?
• how would you solve: 12^log12(4)
(1 vote)
• 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
• Prove that log 11 base 3 lies between 1/2 and 1/3
(1 vote)
• I don't think this is true. And here's why.
3^2 = 9 , so therefore log_3(9) = 2
3^3 = 27, so therefore log_3(27) = 3
Since 11 is between 9 and 27, it follows that
log_3(11) is between log_3(9) and log_3(27).
In other words, log_3(11) is between 2 and 3.
Hope this is helpful to you!