Log2(x+1)=log 8(x+1) срочно помогите

Решите задачу:

$\log_2(x+1)=\log_8(x+1), \\ x+1\ \textgreater \ 0, x\ \textgreater \ 1, \\ \log_2(x+1)=\log_{2^3}(x+1), \\ \log_2(x+1)=\frac{1}{3}\log_2(x+1), \\ \log_2(x+1)=\log_2(x+1)^\frac{1}{3}, \\ x+1=(x+1)^\frac{1}{3}, \\ ((x+1)^\frac{1}{3})^2=1, \\ \left [ {{(x+1)^\frac{1}{3}=-1,} \atop {(x+1)^\frac{1}{3}=1;}} \right. \left [ {{x+1=-1,} \atop {x+1=1;}} \right. \left [ {{x=-2,} \atop {x=0;}} \right. \\ x\in\varnothing.$