<s>[INST] 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 $27$.
 [/INST] Using Henry's approach, but with different estimates. If you know that the first three digits of $\ \sqrt{2}\ $ is $\ 1.41,\ $ then you have: $\ 1.40 < \sqrt{2} < 1.42857\ldots,\ $ i.e.,
$$\ \frac{7}{5} < \sqrt{2} < \frac{10}{7}.$$
So,
$$ \left( 9^{\sqrt{2}} \right)^7 < \left( 9^{ \frac{10}{7}} \right)^7 = 9^{10} = {81}^5 = \left( 80+1 \right)^5 $$
$$ = {80}^5 + 5\times {80}^4 + 10\times {80}^3 + 10\times {80}^2 + 5\times 80 + 1 $$
$$ < {80}^5 + 6 \times {80}^4 = 86 \times {80}^4 < 100 \times{80}^4 $$
$$ =100\times 8^4 \times {10}^4 < 4,100,000,000 = 4.1 \times 10^{9}, $$
whereas
$$ \left(\left(\sqrt{2}\right)^9\right)^7 = \left(\sqrt{2}\right)^{63} = 2^{31}\times \sqrt{2} > 1.4 \times 2^{31} = \frac{7}{2} \times 2^{31} $$
$$ = 7 \times 2^{30} = 7 \times \left( 2^{10} \right)^3 = 7 \times 1024^{3} > 7 \times 1000^3 = 7 \times 10^9. $$
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