Помогите решить , пожалуйста (подробно)

Решите задачу:

$1) log100xlogx=-1 \\ x \geq 0 \\ (log100+logx)logx=-1 \\ (2+logx)logx=-1 \\ log^2x+2logx+1=0 \\ (logx+1)^2=0\iff logx=-1\iff x=10^{-1}=0,1 \\ x=0,1 \\ \\ 2)log(x^2-8)=log(2-9x) \\ a) x^2-8\ \textgreater \ 0\rightarrow (x-2 \sqrt{2})(x+2 \sqrt{2})\ \textgreater \ 0\\ \rightarrow x\in(-\infty;-2 \sqrt{2})\vee (2 \sqrt{2};+\infty) \\ b)2-9x\ \textgreater \ 0\rightarrow x\ \textless \ \frac{2}{9} \\ \\ a;b) x\in(-\infty;-2 \sqrt{2}) \\ x^2-8=2-9x\rightarrow x^2+9x-10=0 \\ \Delta=81+40=121 \\ x_1= \frac{-9-11}{2}=-10 \in(-\infty;-2 \sqrt{2}) \\$
$x_2= \frac{-9+11}{2}=1\notin (-\infty;-2 \sqrt{2}) \\$