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?
 A: 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.
====
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.
A: Note: $7^2<2\cdot 5^2$ and $5>2^2$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{55}\cdot 2^{28}<5^{69}<5^{72}$ as required

With an extra jink into factors of $3$, we can show $7^{55}<5^{67}$ 
Extra notes: $3^3>5^2$ and $5^5>3\cdot2^{10}$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{54}\cdot 2^{30}<5^{52}\cdot 2^{30}\cdot 3^{3}<5^{67}$
A: Consider that
$$
\frac{7^{55}}{5^{72}} = 7\left(\frac{7^3}{5^4}\right)^{18}
$$
Now, hand-calculation is easy enough on the bracketed term: $7^3=7\times49=343$ and $5^4=25^2=625$. As such, the bracketed term is just a little larger than $1/2$, and certainly less than $1/\!\sqrt{2}$. Therefore,
$$
7\left(\frac{7^3}{5^4}\right)^{18}<7\left(\frac12\right)^9 = \frac{7}{512}<1
$$
A: We have $7^4 = 49^2 < 50^2 = 4 \times 5^4 < 5^5$. Hence 
$$7^{55} < 7^{56} = (7^4)^{14} < (5^5)^{14} = 5^{70} < 5^{72}.$$
A: Observe that $7^2=49<2\cdot 5^2$.  In this case,
$$
7^{55}=7\cdot 7^{54}=7\cdot(7^2)^{27}<7\cdot (2\cdot 5^2)^{27}=7\cdot 2^{27}5^{54}.
$$
Observe that $2^3<10=2\cdot 5$.  In this case,
$$
7\cdot 2^{27}5^{54}=7\cdot (2^3)^95^{54}<7\cdot(2\cdot 5)^95^{54}=7\cdot 2^9\cdot 5^{63}
$$
Using that $2^3<10=2\cdot 5$ again, we get
$$
7\cdot 2^9\cdot 5^{63}=7\cdot (2^3)^3\cdot 5^{63}<7\cdot (2\cdot 5)^3\cdot 5^{63}=7\cdot 2^3\cdot 5^{66}.
$$
Since $7\cdot 2^3=7\cdot 8=56<125=5^3$, we get
$$
7\cdot 2^3\cdot 5^{66}<5^{69}<5^{72}.
$$
A: Observe that $7^2/5^2 = 49/25 < 2$.  Cubing both sides, $(7^2/5^2)^3 < 8$.  Since $7/5^3 = 7/125 < 1/8$ we have $7^7/5^9 < 8 \times 1/8 = 1$.
Then $7^{55}/5^{72} = 1/7 \times (7^7/5^9)^8 < 1/7  < 1$ and so $7^{55} < 5^{72}$.
