Show that $1/2^{100\log(n)}$ and $e^{-100\log(2) \log(n)}$ are equal

Question

Suppose $A = 1/2^{100\log(n)}$, and $B = e^{-100\log(2) \log(n)}$.

I'm required to prove that $A$ and $B$ are equal, how should I prove this? I tried applying some rules of logarithms that I have learned but I'm not able to show this.

Answer 1

hint

take logarithm and use

$$\ln(A)=\ln(B) \implies A=B$$

and

$$\ln(\frac 12)=-\ln(2)$$