Помогите решить уравнение с логарифмом в степени

$2^{log_{16}(6x+7)}=7$
$2^{log_{2^{4}}(6x+7)}=7$
$2^{ \frac{1}{4}*log_{2}(6x+7)}=7$
$2^{log_{2}( \sqrt[4]{6x+7}) }=7$
$\sqrt[4]{6x+7}=7$

0} \atop {6x+7=7^{4}}} \right. " alt=" \left \{ {{6x+7>0} \atop {6x+7=7^{4}}} \right. " align="absmiddle" class="latex-formula">

-\frac{7}{6} } \atop {6x=7^{4}-7}} \right. " alt=" \left \{ {{x>-\frac{7}{6} } \atop {6x=7^{4}-7}} \right. " align="absmiddle" class="latex-formula">

-\frac{7}{6} } \atop {x=\frac{7}{6}*(7^{3}-1)}} \right. " alt=" \left \{ {{x>-\frac{7}{6} } \atop {x=\frac{7}{6}*(7^{3}-1)}} \right. " align="absmiddle" class="latex-formula">

$x=\frac{7*342}{6}=7*57=399$ - ответ

Проверка:
$2^{log_{16}(6*399+7)}=2^{log_{16}(2401)}=2^{log_{2}(\sqrt[4]{2401})}=\sqrt[4]{2401}=7$ - верно

Ответ: x=399