Thread: Solving for X in both sides

1. 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.

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

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

Can you take it from here?

3. 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.

4. 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:

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

Isolate x:

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

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

And now divide...

5. Thank you alot. That helped a lot. I'm glad I learn best by example

6. I worked out 1.51..... is that correct?

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