# 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?

• If you're not going to use a calculator, then you have to try to approximate the two powers. For example, $7^2\approx 2*5^2$. Really, logarithms are the best way to solve this type of question. Feb 23, 2017 at 17:18
• Purely heuristic argument: Exponential functions have a few interesting properties: 1) they grow really, really fast, 2) when the bases are of a similar scale, the functions exhibit similar growth rates. So an increase of 17 in the exponent is huge relative to a (multiplicative) increase of 7/5 in the base. This isn't guaranteed to give you the right answer, but it can give you a nice gut feeling for this sort of problem without having to do any "real" math. Feb 24, 2017 at 7:26
• Possible duplicate of Which of these numbers is the biggest Feb 24, 2017 at 7:55

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 $5^5>3\cdot2^{10}$

$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{54}\cdot 2^{30}<5^{52}\cdot 2^{30}\cdot 3^{3}<5^{67}$

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 $$7\cdot 2^3\cdot 5^{66}<5^{69}<5^{72}.$$

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.

• Typo alert: line 6. 49 / 5 should be 49 / 50. Feb 23, 2017 at 23:33

We have $7^4 = 49^2 < 50^2 = 4 \times 5^4 < 5^5$. Hence $$7^{55} < 7^{56} = (7^4)^{14} < (5^5)^{14} = 5^{70} < 5^{72}.$$

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, $$7\left(\frac{7^3}{5^4}\right)^{18}<7\left(\frac12\right)^9 = \frac{7}{512}<1$$

• $7 \times 49 = 343$, but this solution still works. Feb 24, 2017 at 14:54
• @MichaelLugo - oops, minor mistake when doing it in my head. Fixed. Feb 24, 2017 at 16:33

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 $7^{55} < 5^{72}$.