How do you evaluate ${log}_{32} 2$ $log_32 2$ ?

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Answer 1 · 2016-08-08T15:33:06

$\frac{1}{5}$ $1/5$

By the logarithm definition,

$y = {log}_{32} 2 \equiv 32^{y} = 2$ $y = log_32 2 equiv 32^y = 2$ but

$32^{y} = {(2^{5})}^{y} = 2^{5 y} = 2 = 2^{1}$ $32^y = (2^5)^y = 2^{5y} = 2 = 2^1$

concluding $5 y = 1$ $5y=1$ then $y = \frac{1}{5}$ $y = 1/5$

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