Calculus - Logarithms

[MohtasaUnique]

So I just started my calc college course this semester, and I'm already lost because I missed a day of lecture (dentist)

The unit we're in is logarithms and I know how to do all the basic stuff like go from logs to exponents and back again, and simplifying log equations and all that, but I don't know how to solve for x on more complex log equations.

So an example of where I'm stuck is this:

3^(x-1) = 2^(2x-1)

It's probably something simple that I'm not getting, but this is how far I can go before getting stuck:

3^(x-1) = 2^(2x-1)
log3^(x-1) = log2^(2x-1)
(x-1)log3 = (2x-1)log2

And I don't know where to go from there. The answer in the book is
x = (log3 - log2)/(log3 - 2log2)
But it doesn't show how to get there. I'm sooooooo confused and the test is tuesday :[

 

[Rsaccounttrader]

(x-1)log3=2x-1)log2
log3/log2=2x-1/x-1
Do log3/log2 on calc (lets say for now it equals ANS)
then
ANS=2x-1/x-1
ANSx-ANS=2x-1
ANSx-ANS+1=2x
1-ANS=2x-ANSx
1-ANS=x(2-ANS)
x=(1-ANS)/(2-ANS)

 

[MohtasaUnique]

Alright, time for another equation: Logₓ1/8 = -1/3. The back of the book says x = 512, and every working online calculator says the same, but I can't get the result.

So far I've got:
Loxₓ1/8 = -1/3 --> X^(-1/3) = 1/8

But I'm lost as to what to do next.

 

[The Fat Controller]

(x^(-1/3))^3 = (1/8)^3 (cube both sides)
x^-1 = 1/x = 1/512
x = 512 (flip the fractions)