Question

8(1/3+log2(3) / log2log3(81);

Answer 1

$dispaystyle frac{8^{1/3+log_23}}{log_2log_381}= frac{(2^3)^{1/3+log_23}}{log_2log_33^4}= frac{2^{1+3log_23)}}{log_24}= frac{2*2^{log_23^3}}{log_22^2}=\= frac{2*3^3}{2}=3^3=27$