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

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Answer 1 · 2017-01-15T12:33:36

$x = {log}_{3} 5$ $x=log_3 5$

First you would divide all the terms by 4:

$3^{x} = 5$ $3^x=5$

then, through the equivalence:

$b^{x} = a \Leftrightarrow x = {log}_{b} a$ $b^x=ahArrx=log_ba$ ,

you get

$x = {log}_{3} 5$ $x=log_3 5$

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