# Logarithms help

• July 8th 2009, 01:09 PM
greghunter
Logarithms help
Solve $2^{x+1}=3^x$.
I've done quite a bit of logs but am not sure how to work this one out.
Thanks for any help.
• July 8th 2009, 01:36 PM
AlephZero
Just take the bit-log of each side:

$2^{x+1}=3^x$
$\implies \log_2(2^{x+1})=\log_2 3^x$
$\implies x+1=x\log_2 3$
$\implies x(1-\log_2 3)=-1$
$\implies x=\frac{-1}{1-\log_2 3}$

which I believe is equivalent to $x=\frac{-\ln 2}{\ln 2 - \ln 3}$, if you prefer.