\(2^{\frac{1}{3}} \log_2 64=\sqrt [3]{2} \log_2 2^6 \)
we know \( \log_a x^n=n \log_a x \) and \( \log_n n=1 \) , so we have:
\(2^{\frac{1}{3}} \log_2 64=\sqrt [3]{2} \times 6 \log_2 2=\sqrt [3]{2} \times 6 \)
and placement \( \sqrt[3]{2}=1.25992 \):
\(2^{\frac{1}{3}} \log_2 64=1.25992 \times 6=7.55952 \)
