Simple Logarithms Question

Hi all,

I have returned to maths after a break of five years. Although I was quite good at it in school, I studied a non-technical subject at university and I have lost nearly all my mathematical ability.

I encountered this problem while attempting to revise logarithms:

Use logarithms to solve the following equation:

$\displaystyle 10^x = e^{2x-1}$

I attempted to solve it as follows:

$\displaystyle x \log 10 = 2 x \log e - x \log e$

$\displaystyle \log e = 2 x \log e - x \log e$

$\displaystyle \log e = x \log ( \frac {e^2}{10} )$

$\displaystyle x = \frac {\log e}{\log \frac {e^2}{10}}$

$\displaystyle x = \frac {0.43429}{-0.13141}$

$\displaystyle x = -13.82649$

This is not the correct answer. I would be grateful if somebody told me where I am going wrong.

Regards,

Evanator