<s>[INST] Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
 [/INST] Okay you have $A = \frac {7^{55}}{5^{72}}=(\frac 75)^{55}\times(\frac 15)^{17}$
well, since you choose to go that way:
$= (\frac {49}{25})^{27}(\frac 1{5})^7\times [\frac 7{5}]$
$= (2*\frac{49}{50})^{27}(\frac 14\times \frac45)^{17}[\frac 7{5}]$
$=2^{27}\times2^{-34}\times[(\frac{49}{50})^{27}\times (\frac 45)^{17}\times \frac 75]$
As $\frac {49}{50} < 1$ and $\frac 45 < 1$ then
$< 2^{27}\times2^{-34}\times \frac 75$
$= \frac 7{2^7*5} < 1$.
So $7^{55} < 5^{72}$.  By quite a lot actually.
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Another way of doing it would be
$\log 7^{55} = 55 \log 7$ and $\log 5^{72} = 72\log 5$
And $\log 7 = \frac 12 \log 49 \approx^- \frac 12 \log \frac{100}2 = 1 - \frac {\log 2}2$.  So $\log 7^{55} \approx^- 55 - 22\frac 12 \log 2$
$\log 5 = \log \frac {10}2 = 1 - \log2$. So $\log^{72} = 72 - 72\log 2$
So $7^{55} ??? 5^{72}$
if $55 - 22\frac 12 \log 2 ???^- 72 - 72\log 2$
$49\frac 12 \log 2 ???^- 17$
And $\log 2 = \frac 1{10} \log 2^{10} = \frac 1{10} \log 1024 \approx^+ \frac 3{10}$
So $49\frac 12 \log 2 \approx^+  4.95\times 3 < 17$.  
There is a bit of margin of error as $\log 2 > 3/10$ and $\log 7 < 1 - \frac {\log 2} 2$ but the margin is not significant.
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