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# 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?

Note by Muzaffar Ahmed
3 years, 4 months ago

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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.

- 3 years, 4 months ago

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

- 3 years, 4 months ago

\begin{align} \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{align} Therefore \begin{align} 19 \log 17 &> 17\log 19 \\ 17^{19} &> 19^{17} \end{align}

Staff - 3 years, 4 months 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 $$x > e$$ ?

- 3 years, 4 months 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 $$18^{19} > 19^{18}$$.

Staff - 3 years, 4 months ago

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

- 3 years, 4 months 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 $$17^{19}>18^{18}>19^{17}$$

Staff - 3 years, 4 months ago

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

- 3 years, 4 months ago

Absolutly cz 17 is smaller than 19

- 3 years, 4 months ago

No

- 3 years, 4 months ago