# Solving for X in both sides

• Oct 12th 2009, 01:38 PM
Velvet Love
Solving for X in both sides
3^(7x-1) = 5^(3x+2)

I've tried to take the natural log of both sides, and pull x out of the exponent, but I can't seem to get the right answer.
• Oct 12th 2009, 01:42 PM
VonNemo19
Quote:

Originally Posted by Velvet Love
3^(7x-1) = 5^(3x+2)

I've tried to take the natural log of both sides, and pull x out of the exponent, but I can't seem to get the right answer.

$\displaystyle \ln{3^{(7x-1)}}=\ln{5^{(3x+2)}}$

$\displaystyle \Rightarrow(7x-1)\ln{3}=(3x+2)\ln{5}$

Can you take it from here?
• Oct 12th 2009, 01:47 PM
Velvet Love
That's what I've gotten it down to. Then I end up with ((ln(5)/ln(3))+3)/4, which is wrong. I'm rusty at pre-cal.
• Oct 12th 2009, 01:57 PM
VonNemo19
Quote:

Originally Posted by Velvet Love
That's what I've gotten it down to. Then I end up with ((ln(5)/ln(3))+3)/4, which is wrong. I'm rusty at pre-cal.

Distribute:

$\displaystyle (7\ln3)x-\ln{3}=(3\ln{5})x+2\ln{5}$

Isolate x:

$\displaystyle (7\ln3)x-(3\ln{5})x=2\ln{5}+\ln3$

$\displaystyle x(7\ln{3}-3\ln{5})=2\ln{5}+\ln3$

And now divide...
• Oct 12th 2009, 02:12 PM
Velvet Love
Thank you alot. That helped a lot. I'm glad I learn best by example :D
• Oct 15th 2009, 09:16 AM
sanddollar
I worked out 1.51..... is that correct?
• Oct 15th 2009, 09:46 AM
BabyMilo
Quote:

Originally Posted by sanddollar
I worked out 1.51..... is that correct?