# Challenge 23: Fractional exponents

1. Nov 17, 2014

### Greg Bernhardt

With only only paper & pencil (no calculator or logarithmic tables), figure out which of the following expressions has a greater value: 101/10 or 31/3.

Please make use of the spoiler tag and write out your full explanation, not just the answer.

Last edited: Nov 17, 2014
2. Nov 17, 2014

### Joffan

Set $x = 10^{1/10}$
Set $y=3^{1/3}$

$x^{30} = 10^3 = 1000$
$y^{30} = 3^{10} = 9^5 > 81^2 > 6400$

$3^{1/3} > 10^{1/10}$

3. Nov 17, 2014

### Greg Bernhardt

I see that was still too easy :)

4. Nov 17, 2014

### Joffan

Also...
for $y > x > e$, $x^y > y^x$, since $\frac{\ln x}{x}$ reaches maximum at $e$.

5. Nov 17, 2014

### ZetaOfThree

Make a guess that $3^{1/3}>10^{1/10}$. This is true iff $3^{10} > 10^3 = 1000$. We can see that this is true since $3^{10} = 27^{3} \cdot 3>27^3 > 10^3$. Therefore we made the right guess that $3^{1/3}>10^{1/10}$.

Last edited: Nov 17, 2014
6. Nov 26, 2014

### Curious3141

The plodding, rigorous way.

First sketch the curve $y = x^{\frac{1}{x}}$ for real, positive $x$.

It can be shown that the function is always positive, starting from the origin, reaching a maximum of $e^{\frac{1}{e}}$ at $x=e$ then decreasing asymptotically to $1$ as $x \to \infty$. All this can be shown by implicit differentiation and L' Hopital's Rule. There are no other turning points.

Since $3$ and $10$ are both greater than $e$ and the function is decreasing over this interval, that allows us to conclude that $3^{\frac{1}{3}} > 10^{\frac{1}{10}}$.

Taking the $30$th (which is the lcm of $3$ and $10$) power is the quick and elementary way, but this is more general.