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One friend asked me to find which one is bigger: $9^{17}$ and $7^{19}$ using basic calculations only. I gave him a solution by using the technique given in here. However, it was not that basic since I had to go up to $6$th term: ${17\choose 5} \left(\frac{2}{7}\right)^5$, computation of which was not easy without calculator.

Can anyone give me a simpler solution (which does not require calculator)?

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5 Answers 5

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$$7^{19}=7\cdot49^9\lt7\cdot50^9=7\cdot125^3\cdot10^9\lt72\cdot128^3\cdot10^8=9\cdot2^{24}\cdot10^8=9\cdot80^8\lt9\cdot81^8=9^{17}$$

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    $\begingroup$ May I ask how did you come up with these inequalities? Thanks! $\endgroup$
    – awllower
    Commented May 25, 2016 at 8:24
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    $\begingroup$ @awllower: I started from both ends, I saw that both squares differ by just one from a multiple of $10$, so that left $70\cdot5^9\le9\cdot8^8$ to prove, then I realised that $5^3\approx2^7$, and that left $70\lt72$. $\endgroup$
    – joriki
    Commented May 25, 2016 at 8:34
  • $\begingroup$ @joriki Bravo Sir, bravo! $\endgroup$
    – user284001
    Commented May 25, 2016 at 8:41
  • $\begingroup$ Thanks for the nice explanations! $\endgroup$
    – awllower
    Commented May 25, 2016 at 9:35
  • $\begingroup$ Really nice! I've added a link to it in an answer to a related question, math.stackexchange.com/questions/1788290/… $\endgroup$
    – Barry Cipra
    Commented May 25, 2016 at 21:45
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An alternative approach, assuming this is doable enough by hand: $$\color{blue}{\left(\frac{7}{9}\right)^3} = \frac{343}{729} \color{blue}{< \frac{1}{2}}$$ Then: $$\color{red}{\frac{7^{19}}{9^{17}}} = \left(\frac{7}{9}\right)^{18} \; \frac{9}{7} \; 7^2 = \left( \color{blue}{\left(\frac{7}{9}\right)^3} \right)^6 \; 63 \color{blue}{<} \left(\color{blue}{ \frac{1}{2}} \right)^6 63 = \frac{63}{64} \color{red}{< 1}$$ So $\color{red}{7^{19}<9^{17}}$ after fairly easy and straightforward calculations.

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Hint: the following are equivalent: $$ 9^{17} > 7^{19}\\ 17 \log 9 > 19 \log 7\\ \frac{\log 9}{10+9} > \frac{\log 7}{10+7} $$

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    $\begingroup$ The inequiality is not obvious IMO. LHS numerator is larger than RHS numerator, and LHS denominator is larger than RHS denominator. If you meant that one could use a calculator at this point, then one could also use a calculator for the original expressions. $\endgroup$
    – barak manos
    Commented May 25, 2016 at 8:08
  • $\begingroup$ @barakmanos Why don't you consider the function $f(x)=\frac{log x}{10+x}$ and see where it is increasing or decreasing? $\endgroup$
    – Fermat
    Commented May 25, 2016 at 8:12
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    $\begingroup$ @Fermat: Because it has a maximum at approximately $8.6$. $\endgroup$
    – joriki
    Commented May 25, 2016 at 8:12
  • $\begingroup$ @joriki Yes, you are right. $\endgroup$
    – Fermat
    Commented May 25, 2016 at 8:16
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Hint write $9=7\times 1.289$ then $9^{17}=7\times 1.28^{17}$. Now divide it by $7^{19}$ you get $\frac{{1.28}^{17}}{49}$ .Now using binomial for $(1+0.28)^{17}$ and writing terms till $0.28^4$ we get value approximately as $66$ so $9^{17}>7^{19}$ note that according to me there will be always some calculation as it's a question on number theory.

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Consider $R=\frac {19^{17}}{17^{19}}$

$R=\frac {19^{17}}{17^{17}17^{2}}=\frac 1{289} \frac {19^{17}}{17^{17}}$

$R=\frac 1{289}(1+ \frac {2}{17})^{17}$

It is clear that $\frac {2}{17}=\frac {6}{51}<\frac {6}{50} \Rightarrow (1+\frac {2}{17})<(1+\frac {6}{50})$

So $R<\frac 1{289}(1+ \frac {3}{25})^{17}$

$R<\frac 1{289}(1 + 17 \frac {3}{25} + {{16.17} \over 2} \frac {3}{25}\frac {3}{25}+{{15.16.17} \over {2.3}} \frac {3}{25}\frac {3}{25}\frac {3}{25} + ... )$

$R<\frac 1{289}(1 + \frac {17.3}{25} + {{17.3} \over 25} (\frac {16}{2}\frac {3}{25})+{{17.3} \over 25} (\frac {16}{2}\frac {3}{25})(\frac {15}{3}\frac {3}{25})+{{17.3} \over 25} (\frac {16}{2}\frac {3}{25})(\frac {15}{3}\frac {3}{25})(\frac {14}{4}\frac {3}{25}) + ... )$

Each successive term is now being multiplied by something like $(\frac {18-k}{k}\frac {3}{25})<(\frac {15}{3}\frac {3}{25})=\frac {3}{5}$

So $R<\frac 1{289}(1 + \frac {17.3}{25} + {{17.3} \over 25} (\frac {16}{2}\frac {3}{25})+{{17.3} \over 25} (\frac {16}{2}\frac {3}{25})(\frac {3}{5})+{{17.3} \over 25} (\frac {16}{2}\frac {3}{25})(\frac {3}{5})^2 + {{17.3} \over 25} (\frac {16}{2}\frac {3}{25})(\frac {3}{5})^3... )$

Sum of infinite geometric sequence gives us

So $R<\frac 1{289}(1 + \frac {17.3}{25} + {{17.3} \over 25} (\frac {16}{2}\frac {3}{25})(\frac {5}{2}) )$

So $R<\frac {7.936}{289}$

Thus $R<1$

$\frac {19^{17}}{17^{19}}<1$

${19^{17}}<{17^{19}}$

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    $\begingroup$ Please check the question. :-) $\endgroup$
    – hola
    Commented May 26, 2016 at 7:29
  • $\begingroup$ Oops! Will fix that. $\endgroup$
    – tomi
    Commented May 26, 2016 at 7:51

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