How to 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

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

Answer 2

Got it now!

$e^{\log x}=x$, so $e^{\log A} = A$.

Consider now, \begin{align*} A &= e^{\log A}\\ &= e^{\log{(1/2^{100\log(n)})}}\\ &=e^{-\log{2^{100\log(n)}}} \quad &&\text{(division rule)}\\ &=e^{-100\log(2)\log(n)} \quad &&\text{(power rule)}\\ &=B. \end{align*}

Thank you for the hints.