17 or 19?

Does anyone have a good and short proof for 1719>1917 17^{19} > 19^{17} without using any help from calculators or log tables?

Note by Muzaffar Ahmed
5 years, 5 months ago

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log1719=19log17=(171917)(log19log17log19)=(log1717log1919)>1 because logxx is strictly decreasing for x>e17log19\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 19log17>17log191719>1917\begin{aligned} 19 \log 17 &> 17\log 19 \\ 17^{19} &> 19^{17} \end{aligned}

Josh Silverman Staff - 5 years, 5 months ago

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

Muzaffar Ahmed - 5 years, 5 months ago

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Consider the function f(x)=x1xf(x) = \displaystyle x^{\frac{1}{x}}. If you test it's derivatives, you'll find it assumes it's maximum at x=ex=e. So ff is larger for that number which is closer to ee as f<0f' <0 for x>ex>e. Particularly here, e<17<19e< 17 < 19, so 17117>191191719>191717^{\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 - 5 years, 5 months ago

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

Nishant Sharma - 5 years, 5 months ago

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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 n3n\geq 3, we have nn+1>(n+1)nn^{n+1}>(n+1)^n . ( Do you see why you need n greater than 3?).

Step 2: compare 1718>181717^{18}> 18^{17} and 1819>191818^{19} > 19^{18} .

Calvin Lin Staff - 5 years, 5 months ago

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

Muzaffar Ahmed - 5 years, 5 months ago

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Show that for n3n \geq 3, we have n1n+1 n^{ \frac 1 {n+1}} is a decreasing sequence and that n1n1 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 1719>1818>191717^{19}>18^{18}>19^{17}

Calvin Lin Staff - 5 years, 5 months ago

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

Siva Prasad - 5 years, 5 months ago

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No

Parth Zain - 5 years, 5 months ago

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Absolutly cz 17 is smaller than 19

Ali Shkeir - 5 years, 5 months ago

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