How to find out the greater number from $15^{1/20}$ and $20^{1/15}$? I have two numbers $15^{\frac{1}{20}}$ & $20^{\frac{1}{15}}$. 
How to find out the greater number out of above two?
I am in 12th grade. Thanks for help!
 A: Well raise both numbers to the power of $20$
That is 
$$\large{(15^\frac{1}{20})^{20} = 15^\frac{20}{20} = 15}$$
Now $$\large{(20^\frac{1}{15})^{20} = 20^\frac{20}{15} = 20^\frac{4}{3} = 20^{1.333..}}$$
which is greater ? $\large{15}$ or $\large{20^{1.333...}}$
Clearly , it is $\large{20^{1.333..}}$ because $\large{20^{1.333} > 20^1 > 15}$ and so this means that $\large{20^\frac{1}{15} >15^\frac{1}{20}}$
A: $15<20, 15^{1/20}<20^{1/20}$  as $\dfrac1{20}>0$
Now, $20^{1/20}<20^{1/15}$  as $\dfrac1{20}<\dfrac1{15}$

Alternatively, $$15^{1/20}<=>20^{1/15}\iff15^{15}<=>20^{20}$$
Now $15^{15}<20^{15}<20^{20}$
A: This is just another way of looking at alkabary's proof.
\begin{align}
   \left[ \dfrac{20^{\frac{1}{15}}}
         {15^{\frac{1}{20}}} \right]^{20}
   &=\dfrac{20^{\frac 43}}{15} \\
   &= \dfrac 43 20^{\frac 13} \\
   &> \dfrac 43 \cdot 2 \\
   &> 1
\end{align}
$\left[ \dfrac{20^{\frac{1}{15}}}
         {15^{\frac{1}{20}}} \right]^{20} > 1
 \implies \dfrac{20^{\frac{1}{15}}} {15^{\frac{1}{20}}} > 1
 \implies 20^{\frac{1}{15}} > 15^{\frac{1}{20}}$
