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. 

Want to prove that  


*

*$\log_5{7} = \frac{\lg{7}}{\lg{5}} < \sqrt{2}$


Equivalently we can show that 


*

*$\lg{7} < \lg{5}\times\sqrt{2}$  

*$7 < 5^{\sqrt{2}}$


where $\lg$ is the base 2 logarithm. Notice that  


*

*$5\times5^{\frac{2}{5}}= 5^{1.4} <5^{\sqrt{2}}$


So can we show that $\frac{7}{5} < 5^{\frac{2}{5}}$? Sure, since $7<8=32768^{\frac{1}{5}}<78125^{\frac{1}{5}}$. Hence  


*

*$7 < 5^{1.4} <5^{\sqrt{2}}$

