Mathematics

The key is to compare these irrational exponents by taking logarithms, which converts the comparison into one about rational numbers. Taking $\ln$ of both sides transforms $3^{\sqrt{2}} > 2^{\sqrt{3}}$ into $\sqrt{2} \ln 3 > \sqrt{3} \ln 2$, which you can then rearrange to $\frac{\ln 3}{\sqrt{3}} > \frac{\ln 2}{\sqrt{2}}$—a much more concrete inequality to verify numerically or algebraically.

The key is to compare these irrational exponents by taking logarithms, which converts the comparison into one about rational numbers. Taking $\ln$ of both sides transforms $3^{\sqrt{2}} > 2^{\sqrt{3}}$ into $\sqrt{2} \ln 3 > \sqrt{3} \ln 2$, which you can then rearrange to $\frac{\ln 3}{\sqrt{3}} > \frac{\ln 2}{\sqrt{2}}$—a much more concrete inequality to verify numerically or algebraically.