How do you solve $4(3^x)=20$ ?

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Answer 1 · 2017-01-15T12:33:36

$x=log_3 5$

First you would divide all the terms by 4:

$3^x=5$

then, through the equivalence:

$b^x=ahArrx=log_ba$ ,

you get

$x=log_3 5$

How do you solve 4(3^x)=204(3^x)=20?