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. 
 A: 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.
A: $f(x)=x^\frac1x$ is a function defined on $(0,\infty)$
its log is $$G(x)=\log f(x)=\frac{\log x}x$$
$$G'(x)=\frac{1-\log x}{x^2}$$
Therefore $G(x)=\log f(x)$ strictly decreases for $x>e$, but logarithm is monotone on $(0,\infty)$ so that $f(x)$ is strictly decreasing for $x>e$
This gives us $$5^{\frac15}>7^{\frac17}$$ implying (by taking 7th power) that $$5^\sqrt2>5^{1.4}>7$$
A: 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  


*

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

A: Define $f(x)=5^{x/5}-x$.
If $x>5$ is obvious that $f'(x)=5^{(-1+x/5)}\ln (5)-1>0.$
Since $f(5)=0$, we have $f(7)>0$.
Since $\frac{7}{5}=\sqrt{\frac{49}{25}}<\sqrt{2}$ we are done.
A: Prove that $\log _5 7 < \sqrt 2$
Lemma
Mathematicians use the word "lemma" to refer to a fact about math which helps us prove other facts about math.
The following is a useful lemma:

Let $x$ and $y$ be any two positive decimal numbers.
For example,

*

*$x$ could be $\pi \approx 3.141592653589793238462643383282 \cdots$

*$x$ could be $\frac{4}{3}$

*$x$ could be the number $9$

*$x$ could be $0.407129911$

  If
      there exists a positive whole number $k$ such that $\lfloor{x * 10^{k}}\rfloor > \lfloor{y * 10^{k}}\rfloor$
  then
     $x > y$

If you are unfamiliar with the notation $\lfloor{x}\rfloor$, it means "delete everything to the right of the decimal point."
For example,

*

*$\lfloor{3.1459}\rfloor = 3$

*$\lfloor{45.28}\rfloor = 45$

*$\lfloor{1.5}\rfloor = 1$

In other words, if you multiply both decimal numbers by ten a few times, and then throw away everything to the right of the decimal point, then you can sometimes tell which number is bigger.

$ \begin{align}
log_5(7) & \approx 1.209061 \dots \\
 \sqrt{2} & \approx 1.4142136 \dots \\
\end{align}  $

$\begin{align}
\lfloor 10*\sqrt{2} \rfloor & = 14   \\
\lfloor 10*\log_5(7)\rfloor & = 12 \\
\end{align} $
$14 = \lfloor 10*\sqrt{2} \rfloor > \lfloor 10*\log_5(7) \rfloor = 12$
Therefore, $\sqrt{2} > log_5{7}$

