7

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)?

Community
  • 1
hola
  • 1,339

5 Answers5

25

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

joriki
  • 238,052
  • 2
    May I ask how did you come up with these inequalities? Thanks! – awllower May 25 '16 at 08:24
  • 4
    @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 $70\lt72$. – joriki May 25 '16 at 08:34
  • @joriki Bravo Sir, bravo! –  May 25 '16 at 08:41
  • Thanks for the nice explanations! – awllower May 25 '16 at 09:35
  • Really nice! I've added a link to it in an answer to a related question, http://math.stackexchange.com/questions/1788290/which-of-the-numbers-is-larger-794-or-991/1789159#1789159 – Barry Cipra May 25 '16 at 21:45
13

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.

StackTD
  • 27,903
  • 34
  • 63
0

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

mookid
  • 28,236
  • 6
    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. – barak manos May 25 '16 at 08:08
  • @barakmanos Why don't you consider the function $f(x)=\frac{log x}{10+x}$ and see where it is increasing or decreasing? – Fermat May 25 '16 at 08:12
  • 3
    @Fermat: Because it has a maximum at approximately $8.6$. – joriki May 25 '16 at 08:12
  • @joriki Yes, you are right. – Fermat May 25 '16 at 08:16
0

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.

Fermat
  • 5,230
Archis Welankar
  • 15,910
0

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 $R<1$

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

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

tomi
  • 9,594