Prove that $\log _5 7 < \sqrt 2.$ 
Prove that $\log _5 7 < \sqrt 2.$

Trial : Here $\log _5 7 < \sqrt 2 \implies 5^\sqrt 2  <7.$ But I don't know how to prove this. Please help. 

Observe that:
$$ \begin{align*}
\log_5 7 &= \dfrac{3}{3}\log_5 7 \\
&= \dfrac{1}{3}\log_5 7^3 \\
&= \dfrac{1}{3}\log_5 343 \\
&< \dfrac{1}{3}\log_5 625\\
&= \dfrac{1}{3}\log_5 5^4\\
&= \dfrac{1}{3}(4)\\
&= \sqrt{\dfrac{16}{9}}\\
&< \sqrt{\dfrac{18}{9}}\\
&= \sqrt{2}\\
\end{align*} $$
as desired.
