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.
–
Paramjit Singh
·
2 years, 7 months ago

@Josh Silverman
–
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
·
2 years, 7 months ago

Log in to reply

(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
·
2 years, 7 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 \(17^{19}>18^{18}>19^{17}\)
–
Calvin Lin
Staff
·
2 years, 7 months ago

Log in to reply

17^{17+2} or (17+2)^{17} Which is greater...?
–
Siva Prasad
·
2 years, 7 months ago

Log in to reply

Absolutly cz 17 is smaller than 19
–
Ali Shkeir
·
2 years, 7 months ago

## Comments

Sort by:

TopNewestConsider 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. – Paramjit Singh · 2 years, 7 months agoLog in to reply

– Nishant Sharma · 2 years, 7 months ago

That's what we learnt in our school. For our exams... Nice exampleLog in to reply

\[\begin{align} \log 17^{19} &= 19\log 17 \\ &=\left(17\frac{19}{17}\right)\left(\log 19 \frac{\log 17}{\log 19}\right) \\ &= \underbrace{\left(\frac{\frac{\log 17}{17}}{\frac{\log 19}{19}}\right)}_{>1 \text{ because }\frac{\log x}{x}\text{ is strictly decreasing for }x>e }17\log 19 \\ \end{align}\] Therefore \[\begin{align} 19 \log 17 &> 17\log 19 \\ 17^{19} &> 19^{17} \end{align} \] – Josh Silverman Staff · 2 years, 7 months ago

Log in to reply

– Muzaffar Ahmed · 2 years, 7 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 \( x > e \) ?Log in to reply

(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 · 2 years, 7 months ago

Log in to reply

@Calvin Lin :) – Muzaffar Ahmed · 2 years, 7 months ago

Ah, yes, a decent solution.. Thanks BroLog in to reply

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 · 2 years, 7 months ago

Log in to reply

17^{17+2} or (17+2)^{17} Which is greater...? – Siva Prasad · 2 years, 7 months ago

Log in to reply

Absolutly cz 17 is smaller than 19 – Ali Shkeir · 2 years, 7 months ago

Log in to reply

No – Parth Zain · 2 years, 7 months ago

Log in to reply