Question

Simplify log_64 32 = x

Answer 1

@darkknight

Answer 2

Log base 64 of 32?

Answer 3

i think?

Answer 4

So log equations are written in this form:
log x = y
b

Then b^y = x

Answer 5

So in this case Log 32 = x
64

64^x = 32
Can you solve for x?

Answer 6

1/6?

Answer 7

Well no, 64^(1/6) would be 2
But think about what it would be. If you do take the 6th root, then do to the power of 5. It will be the same as 5/6, and that would get you to 32

Answer 8

So solving for x would get us 5/6

Answer 9

I don't think that was a good explanation, so you have 64^x = 32. Are you asking how to solve that?

Answer 10

yes

Answer 11

I dont really get it until i show through work lol

Answer 12

Okay, so how to do this to take the log base of both sides.
So here we have 64^x = 32
so we take \[\log_{64} \] of both sides to cancel out the 64. So x = \[\log_{64}(32) \]

Answer 13

ok so i got that

Answer 14

Alright. So if you ever approach a problem like that and have a variable as the exponent, what you can do is to take the log base of the base. So for ex: If you have b^y = x
And y is your variable, then you can preform this: \[\log_{b}(b)^y = \log_{b}x \] which equals \[y= \log_{b}x \] Which, rewriting using change of base formula, is \[y = \log(x)/\log(b)\]

Answer 15

Hope that helped :)