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.3^(3x-1)=2^(2x-1)
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.
So your exact answer is:
x = (log 3 - log 2) over (3 log 3 - 2 log 2)
Note: To cancel out a term or number. You have to do the opposite sign of what it is telling you to do to bring to the other side. For example. If you want to know all the ways that 2+3 = 5. You have to either subtract 2 or 3 on both sides to cancel it out on the left side to bring it to the right side.