Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.

$(\sqrt{2})^9 \quad vs.\quad 9^\sqrt{2}\qquad$// square both side
$2^9=512 \quad vs.\quad 
3^{4\sqrt{2}} \;≈\; 3^{5.657} \;<\; 3^5\,3^\frac{2}{3}$
$\displaystyle 3^5\,3^\frac{2}{3} = 243×2×\sqrt[3]{1+\frac{1}{8}} \;<\;486×\left(1+\frac{1}{3×8}\right) = 506.25$
$→ (\sqrt{2})^9 \;>\; 9^\sqrt{2}\qquad$
