Thread: 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.

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...



Properties of logs allow us to puill the exponents out, like this


Now solve for x...

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

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:





Now you have to expand them



Now you have to collect like terms



Factor out the greatest common factor



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.

Originally Posted by HallsofIvy

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

I'm working from a different PC as I am currently on vacation. The resolution is terrible. I thought that everything was honky dory.

Thanks.