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.

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- Jul 8th 2009, 12:09 PMgreghunterLogarithms help
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. - Jul 8th 2009, 12:36 PMAlephZero
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.