# Simple Logarithms Question

• Mar 22nd 2010, 09:03 AM
evanator
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
• Mar 22nd 2010, 10:09 AM
onemore
when you log both sides it does not cancel out the e. the only thing that can cancel out an e is a ln.
• Mar 22nd 2010, 10:15 AM
masters
Hi evanator,

Use natural logs.

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

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

$\displaystyle x \ln 10=2x-1$

$\displaystyle 2.302585093 x = 2x-1$

Now, can you finish up?
• Mar 22nd 2010, 10:37 AM
evanator
$\displaystyle 0.302585093x = -1$

$\displaystyle x = - \frac {1}{0.302585093}$

$\displaystyle x = -3.304855$

Thank you both for your help, which I sincerely appreciate. I hope my brain kicks in to gear again soon.

Later,

evanator
• Mar 22nd 2010, 11:49 AM
e^(i*pi)
Quote:

Originally Posted by evanator
$\displaystyle 0.302585093x = -1$

$\displaystyle x = - \frac {1}{0.302585093}$

$\displaystyle x = -3.304855$

Thank you both for your help, which I sincerely appreciate. I hope my brain kicks in to gear again soon.

Later,

evanator

Another way:

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

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

$\displaystyle x \ln (10) = 2x-1$

$\displaystyle 2x - x \ln (10) = 1$

$\displaystyle x(2-\ln(10)) = 1$

$\displaystyle x = \frac{1}{2-\ln(10)}$