Assuming $\displaystyle \log a=\log_{10}a$ , and assuming you made a typo and meant $\displaystyle \log\log x$ in the first exer., then
$\displaystyle \log\log x=3\Longrightarrow 10^3=\log x \Longrightarrow 10^{10^3}=x$
$\displaystyle x^{\log_2x}=32x^4\Longrightarrow \log_2\left(x^{\log_2x}\right)=\log_2\left(32x^4\r ight)\Longrightarrow $ $\displaystyle \log_xx\cdot\log_2x=5+4\log_2x$ ...you finish the argument.
Of course, you must know VERY WELL the definition and properties of logarithms to prove the above.
Tonio