Solve $\displaystyle 2^{x+1}=3^x$.
I've done quite a bit of logs but am not sure how to work this one out.
Thanks for any help.
Just take the bit-log of each side:
$\displaystyle 2^{x+1}=3^x$
$\displaystyle \implies \log_2(2^{x+1})=\log_2 3^x$
$\displaystyle \implies x+1=x\log_2 3$
$\displaystyle \implies x(1-\log_2 3)=-1$
$\displaystyle \implies x=\frac{-1}{1-\log_2 3}$
which I believe is equivalent to $\displaystyle x=\frac{-\ln 2}{\ln 2 - \ln 3}$, if you prefer.