Помогите пожалуйста,ребятки)

Решите задачу:

$\mathtt{(\frac{1}{5})^{x-3}\ \textless \ \frac{1}{25}=(\frac{1}{5})^2,~\to~x-3\ \textgreater \ 2;~x\ \textgreater \ 5}$

$\mathtt{\sqrt{x+16}=x-4;~\left\{{{x+16=(x-4)^2}\atop{x\geq4}}\right\left\{{{x^2-9x=0}\atop{x\geq4}}\right\left\{{{\left[\begin{array}{ccc}\mathtt{x_1=0}\\\mathtt{x_2=9}\end{array}\right}\atop{x\geq4}}\right~OTBET:~x=9}$

$\mathtt{13^{2\log_{13}7}-2=13^{\log_{13}49}-2=49-2=47}$

$\mathtt{\frac{a^5*a^{-8}}{a^{-2}}=a^{5+(-8)-(-2)}=a^{5-8+2}=a^{-1};~6^{-1}=\frac{1}{6}}$

$\mathtt{3^{2x}-4*3^x+3=0;~(3^x-1)(3^x-3)=0;~x(x-1)=0;~x_1x_2=0}$