# Thread: Exponent and Log problem

1. ## Exponent and Log problem

I have this one problem that I'm not sure how to start or end in solving for x.
3^(3x-1)=2^(2x-1)
I have tried taking the log of both sides but it just came out more complicated. Any help would be appreciated.

2. Originally Posted by NDHS
I have this one problem that I'm not sure how to start or end in solving for x.
I have tried taking the log of both sides but it just came out more complicated. Any help would be appreciated.
Taking logs of both sides here will make things much simpler...

$\displaystyle \ln3^{(3x-1)}=\ln2^{(2x-1)}$

Properties of logs allow us to puill the exponents out, like this
$\displaystyle (3x-1)\ln3=(2x-1)\ln2$

Now solve for x...

3. VonNemo19, group things in LaTex with { }. That is, 3^{3x-1} will give you $\displaystyle 3^{3x-1}$.

4. Originally Posted by NDHS
I have this one problem that I'm not sure how to start or end in solving for x.
I have tried taking the log of both sides but it just came out more complicated. Any help would be appreciated.
This is how I do it:

$\displaystyle 3^{(3x-1)}=2^{(2x-1)}$

$\displaystyle (3x-1)log3 = (2x-1)log2$

Now you have to expand them

$\displaystyle 3xlog 3 - log 3 = 2xlog 2 - log 2$

Now you have to collect like terms

$\displaystyle 3x log 3 - 2x log 2 = log 3 - log 2$

Factor out the greatest common factor

$\displaystyle x (3 log 3 - 2 log 2) = log 3 - log 2$

Now bring (3 log 3 - 2 log 2) over to the other side by canceling out so only x remains. You can do this by divinding it on both sides.

VonNemo19, group things in LaTex with { }. That is, 3^{3x-1} will give you $\displaystyle 3^{3x-1}$.