Помогите плиззз)) log3(2x+3)=4 log6(x+7)=log6(6x-13)

ОДЗ:
0;\\2x>-3;\\x> -\frac{3}{2}\\x \in (-\frac{3}{2}; +\infty)" alt="2x+3>0;\\2x>-3;\\x> -\frac{3}{2}\\x \in (-\frac{3}{2}; +\infty)" align="absmiddle" class="latex-formula">
$log_3(2x+3)=4;\\ 2x+3=3^4;\\ 2x+3=81;\\ 2x=81-3;\\2x=78;\\x=39$
Ответ: 39.

ОДЗ: 0;} \atop {6x-13>0;}} \right. \\ \left \{ {{x>-7;} \atop {6x>13;}} \right. \\ \left \{ {{x>-7;} \atop {x> \frac{13}{6} }} \right. \\x \in ( \frac{13}{6}; +\infty) " alt=" \left \{ {{x+7>0;} \atop {6x-13>0;}} \right. \\ \left \{ {{x>-7;} \atop {6x>13;}} \right. \\ \left \{ {{x>-7;} \atop {x> \frac{13}{6} }} \right. \\x \in ( \frac{13}{6}; +\infty) " align="absmiddle" class="latex-formula">
$log_6(x+7)=log_6(6x-13);\\x+7=6x-13;\\x-6x=-13-7;\\-5x=-20;\\x=4$
Ответ: 4.