Решите уравнения log2 (8+7x)= log2 (8+3x)+1

Log₂(8+7x)=log₂(8+3x)+1
ОДЗ:
$\left \{ {{8+7x\ \textgreater \ 0} \atop {8+3x\ \textgreater \ 0}} \right. , \left \{ {{x\ \textgreater \ - \frac{8}{7} } \atop {x\ \textgreater \ - \frac{8}{3} }} \right. =\ \textgreater \ x\ \textgreater \ - \frac{8}{7}$
1=log₂2¹=log₂2

log₂(8+7x)=log₂(8+3x)+log₂2
log₂(8+7x)=log₂((8+3x)*2)
8+7x=16+6x
x=8