# Math Help - [SOLVED] finding x, using the natural logarithm

1. ## [SOLVED] finding x, using the natural logarithm

I am having trouble with this question. I keep getting X to cancel out somehow.

e^(ax) = Ce^(bx), a cannot = b, Find X

2. Originally Posted by dee18
I am having trouble with this question. I keep getting X to cancel out somehow.

e^(ax) = Ce^(bx), a cannot = b, Find X

Take the natural log of both sides.

$\ln(e^{ax}) = \ln(Ce^{bx})$

Using log rules...

$ax\ln(e) = bx\ln(Ce)$

More rules...

$ax(1) = bx[\ln(C)+\ln(e)]$

Finally $ax=bx(\ln(C)+1)$

Can you solve it now?

3. $e^{ax} = Ce^{bx}$

$\Rightarrow C = \frac{e^{ax}}{e^{bx}}$

$\Rightarrow C = e^{(a-b)x}$

Now recall that:

$\ln{A^x} = x\ln{A}$

Solve for x now.

4. Sigh, Chop Suey never fails to do the same problem faster and more elegantly...

5. Thanks guys! I figured it out