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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!

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    $\begingroup$ Try changing the exponents to have common denominators. $\endgroup$ – Joe Moeller Jun 16 '15 at 5:57
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$15<20, 15^{1/20}<20^{1/20}$ 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 $15^{15}<20^{15}<20^{20}$

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  • $\begingroup$ bases are not same $\endgroup$ – Bhaskara-III Jun 16 '15 at 6:00
  • $\begingroup$ @Bhaskara-III, Which line you are pointing to? $\endgroup$ – lab bhattacharjee Jun 16 '15 at 6:00
  • $\begingroup$ I am not able to understand what finally you proved. $\endgroup$ – Bhaskara-III Jun 16 '15 at 6:05
  • $\begingroup$ @Bhaskara-III In the alternative argument, lab raised both sides to the 35th power, which preserves the inequality. Since $20^{20} > 15^{15}$, we may conclude that $20^{\frac{1}{15}} > 15^{\frac{1}{20}}$. $\endgroup$ – N. F. Taussig Jun 16 '15 at 8:43
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    $\begingroup$ @N.F.Taussig, raised both sides $15\cdot20$th power $\endgroup$ – lab bhattacharjee Jun 16 '15 at 8:46
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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 $\large{20^\frac{1}{15} >15^\frac{1}{20}}$

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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}

$\left[ \dfrac{20^{\frac{1}{15}}} {15^{\frac{1}{20}}} \right]^{20} > 1 \implies \dfrac{20^{\frac{1}{15}}} {15^{\frac{1}{20}}} > 1 \implies 20^{\frac{1}{15}} > 15^{\frac{1}{20}}$

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