How do you evaluate $log_32 2$ ?

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Answer 1 · 2016-08-08T15:33:06

$1/5$

By the logarithm definition,

$y = log_32 2 equiv 32^y = 2$ but

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

concluding $5y=1$ then $y = 1/5$

How do you evaluate log_32 2log_32 2?