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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.

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.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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

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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 $x > e$ ?

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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

somecalculation.Log in to reply

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

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(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 $18^{19} > 19^{18}$.

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

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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 $17^{19}>18^{18}>19^{17}$

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

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No

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Absolutly cz 17 is smaller than 19

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