Proof of the equality $\left(\frac{2^{\ln2}}{3^{\ln3}}\right)^{\frac{1}{\ln3-\ln2}}= \frac{1}{6}$

Question

I want to show that $$\left(\frac{2^{\ln2}}{3^{\ln3}}\right)^{\frac{1}{\ln3-\ln2}}= \frac{1}{6}$$

I've tried for a while but I am unable to prove it by only manipulating one side. Any help would be greatly appreciated.

Answer 1

Prove $\frac{2^{\ln2}}{3^{\ln 3}}^{\frac{1}{\ln3-\ln2}}=1/6$

$2^{\ln2}= (e^{\ln 2})^{\ln2 }=e^{(\ln2)^2}$

$\frac{2^{\ln2}}{3^{\ln 3}}^{\frac{1}{\ln3-\ln2}} = [ \frac{e^{(\ln2)^2}}{e^{(\ln3 )^2}}]^{\frac{1}{\ln3-\ln2}}= e^{\frac{(\ln 2)^2-(\ln 3)^2}{\ln 3-\ln2}}$

$=e^\frac{(\ln2-\ln3)(\ln2 + \ln 3)}{\ln 3 - \ln2}=e^{-(\ln2+\ln3)}$

$=e^{-(\ln 3+\ln2)}=e^{-\ln 6}=1/6$