# How to solve 2

How to solve ${2}^{x}=36$
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With a calculator, you can simply calculate:
$x={\mathrm{log}}_{2}36=\mathrm{log}36/\mathrm{log}2$
Without a calculator, you know that x must be a little over 5, since ${2}^{5}=32$. Now:
${2}^{x}={2}^{x-5}{2}^{5}=36\to {2}^{x-5}=36/32=1+1/8$
Using the fact that for small ${\mathrm{log}}_{b}\left(1+x\right)\approx x/\mathrm{ln}b$
$\left(x-5\right)={\mathrm{log}}_{2}\left(1+1/8\right)\approx \frac{1}{8\mathrm{ln}2}$
$x\approx 5+\frac{1}{8\mathrm{ln}2}\approx 5+\frac{1}{5.6}$