How do you solve ${log}_{10} (2^{x}) = {log}_{10} 32$ $log_10 (2^x)=log_10 32$ ?

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Answer 1 · 2016-11-13T20:20:34

$x = 5$ $x=5$

Rewrite $32$ $32$ as a productof $2$ $2$

$32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^{5}$ $32=2*2*2*2*2=2^5$

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

$2^x=2^5$

$x=5$

How do you solve log_10 (2^x)=log_10 32log10(2x)=log1032log_10 (2^x)=log_10 32?