2^(log по основанию 16 от числа (3х-9)) = 3

$2^{log_{16}(3x-9)}=3$

Число 16 - это $2^4$
$2^ {\frac{1}{4}log_{2}(3x-9)}=3$
$2^{log_{2}(3x-9)^{ \frac{1}{4} }}=3$
$(3x-9)^{ \frac{1}{4} }=3$

0}} \right. " alt=" \left \{ {{3x-9=3^{4}} \atop {3x-9 > 0}} \right. " align="absmiddle" class="latex-formula">

3}} \right. " alt=" \left \{ {{3x=3^{4}+9} \atop {x > 3}} \right. " align="absmiddle" class="latex-formula">

3}} \right. " alt=" \left \{ {{x=30} \atop {x> 3}} \right. " align="absmiddle" class="latex-formula">

Ответ: x=30