17 or 19?

Does anyone have a good and short proof for $17^{19} > 19^{17}$ without using any help from calculators or log tables?

#NumberTheory #Exponents #Powers

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

$\begin{aligned} \log 17^{19} &= 19\log 17 \\ &=\left(17\frac{19}{17}\right)\left(\log 19 \frac{\log 17}{\log 19}\right) \\ &= \underbrace{\left(\frac{\frac{\log 17}{17}}{\frac{\log 19}{19}}\right)}_{>1 \text{ because }\frac{\log x}{x}\text{ is strictly decreasing for }x>e }17\log 19 \\ \end{aligned}$ Therefore $\begin{aligned} 19 \log 17 &> 17\log 19 \\ 17^{19} &> 19^{17} \end{aligned}$

Josh Silverman Staff - 7 years ago

Yeah I thought about that.. But.. Is there no other way we can prove that without using the rule of $\frac{log x}{x}$ is decreasing for $x > e$ ?

Muzaffar Ahmed - 7 years ago

Consider the function $f(x) = \displaystyle x^{\frac{1}{x}}$ . If you test it's derivatives, you'll find it assumes it's maximum at $x=e$ . So $f$ is larger for that number which is closer to $e$ as $f' <0$ for $x>e$ . Particularly here, $e< 17 < 19$ , so $17^{\frac{1}{17}} > 19^{\frac{1}{19}} \Rightarrow 17^{19}>19^{17}$ . Sorry bro, calculus is the best option for these problems. You may use a non calculus number theoretic approach but that will be long. Explicitly, you need to do some calculation.

A Brilliant Member - 7 years ago

That's what we learnt in our school. For our exams... Nice example

Nishant Sharma - 7 years ago

(This doesn't work out ... Yet)

A non calculus, non logarithm, non tedious expansion / calculation approach.

Step 1: Show using the binomial theorem that for $n\geq 3$ , we have $n^{n+1}>(n+1)^n$ . ( Do you see why you need n greater than 3?).

Step 2: compare $17^{18}> 18^{17}$ and $18^{19} > 19^{18}$ .

Calvin Lin Staff - 7 years ago

Ah, yes, a decent solution.. Thanks Bro @Calvin Lin :)

Muzaffar Ahmed - 7 years ago

Show that for $n \geq 3$ , we have $n^{ \frac 1 {n+1}}$ is a decreasing sequence and that $n^{ \frac 1 {n-1}}$ is an increasing sequence. The usual approach would be calculus, but you can also use the binomial theorem.

Hence, we have $17^{19}>18^{18}>19^{17}$

Calvin Lin Staff - 7 years ago

17^{17+2} or (17+2)^{17} Which is greater...?

Siva Prasad - 7 years ago

Parth Zain - 7 years ago

Absolutly cz 17 is smaller than 19

Ali Shkeir - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$