Which one is bigger: $9^{17}$ and $7^{19}$ 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)?
 A: $$7^{19}=7\cdot49^9\lt7\cdot50^9=7\cdot125^3\cdot10^9\lt72\cdot128^3\cdot10^8=9\cdot2^{24}\cdot10^8=9\cdot80^8\lt9\cdot81^8=9^{17}$$
A: 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.
A: Hint:
the following are equivalent:
$$
9^{17} > 7^{19}\\
17 \log 9 > 19 \log 7\\
\frac{\log 9}{10+9} > \frac{\log 7}{10+7} 
$$
A: 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.
