2021 AMC 12B Problems/Problem 9

$\frac{log_{2}{80}}{log_{40}{2}}-\frac{log_{2}{160}}{log_{20}{2}}$

Note that $log_{40}{2}=\frac{1}{log_{2}{40}}$ , and similarly $log_{20}{2}=\frac{1}{log_{2}{20}}$

$= log_{2}{80}\cdot log_{2}{40}-log_{2}{160}\cdot log_{2}{20}$

$=(log_{2}{4}+log_{2}{20})(log_{2}{2}+log_{2}{20})-(log_{2}{8}+log_{2}{20})log_{2}{20}$

$=(2+log_{2}{20})(1+log_{2}{20})-(3+log_{2}{20})log_{2}{20}$

Expanding, $2+2log_{2}{20}+log_{2}{20}+(log_{2}{80})^2-3log_{2}{20}-(log_{2}{20})^2$

All the log terms cancel, so answer $=2$ .

~ SoySoy4444

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