How do you solve $log_10 (2^x)=log_10 32$ ?

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Answer 1 · 2016-11-13T20:20:34

$x=5$

Rewrite $32$ as a productof $2$

$32=2*2*2*2*2=2^5$

$:.log_10(2^x)=log_10(2^5)$

$2^x=2^5$

$x=5$

How do you solve log_10 (2^x)=log_10 32log_10 (2^x)=log_10 32?