# How to prove that $7^{31} > 8^{29}$

How can I prove that $7^{31}$ is bigger than $8^{29}$?

I tried to write exponents as multiplication, $2\cdot 15 + 1$, and $2\cdot 14+1$, then to write this inequality as $7^{2\cdot 15}\cdot 7 > 8^{2\cdot 14}\cdot 8$. I also tried to write the right hand side as $\frac{8^{31}}{8^2}$.

• This inequality is equivalent to $31<29 \log_7 8$ or $\frac{31}{29}>\log_7 8$, so if you can find the logarithm and then multiply it by $29$ then you've got it. How to find the logarithm numerically is a substantial question in its own right. ${}\qquad{}$ Jul 7, 2015 at 14:25
• This fails:$$\left( 1 + \frac{1}{7} \right)^{7\cdot 4 \frac{1}{7}} \leq e^{4 \frac{1}{7}} \geq 7^2$$ Any better ideas? Jul 7, 2015 at 14:41
• I think maybe this problem is from my old problem when Christian Blatter solution a step math.stackexchange.com/questions/431461/… Jul 7, 2015 at 16:42
• Maybe you could try to use the binomial theorem. Note that $7=8-1$ and $8=7+1$.
– Surb
Jul 7, 2015 at 16:52
• I'd love to see a proof using the binomial theorem, but it seems that the two values are two close (relatively) for simple bounds to work.
– lhf
Jul 7, 2015 at 16:52

The following (not particularly elegant) proof uses reasonably basic multiplication and division.

We need to show that $7^{31} > 8^{29}$, i.e. that $\dfrac{7^{31}}{8^{29}}>1$.

We have: $\dfrac{7^{31}}{8^{29}}=\dfrac{7^{2}\cdot7^{29}}{8^{29}}=\dfrac{7^{3}}{8}\Big(\dfrac{7}{8}\Big)^{28}=\dfrac{7^{3}}{8}\Big(\dfrac{7^4}{8^4}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{2401}{4096}\Big)^{7} > \dfrac{7^{3}}{8}\Big(\dfrac{2400}{4100}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{24}{41}\Big)^{7}=\dfrac{7^{3}}{8}\dfrac{24}{41}\Big(\dfrac{24}{41}\Big)^{6}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{24^2}{41^2}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1681}\Big)^{3}>\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1683}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{9 \cdot 64} {9\cdot 187}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{64} {187}\Big)^{3}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{11^3 \cdot 17^3 \cdot 41}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{1331 \cdot 17^3 \cdot 41}>\dfrac{2^{18} \cdot 3 \cdot 7^3}{1332 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{111 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{4551 \cdot 17^3}>\dfrac{2^{16} \cdot 7^3}{4557 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{93 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{1581 \cdot 17^2}>\dfrac{2^{16} \cdot 7}{1582 \cdot 17^2}=\dfrac{2^{15}}{113 \cdot 17^2}=\dfrac{2^{15}}{1921 \cdot 17}>\dfrac{2^{15}}{1924 \cdot 17}=\dfrac{2^{13}}{481 \cdot 17}=\dfrac{8192}{ 8177}>1. \quad\square$

• How did you get 7^3/8 * (2500/4100)^7? Jul 7, 2015 at 22:51
• ^^^2400/4100^^^ Jul 7, 2015 at 22:52
• It's an inequality. Jul 7, 2015 at 22:58
• $93\cdot 17^3\equiv 1 \pmod 2$ but $1582\cdot 17^2\equiv 0\pmod 2$. Jul 7, 2015 at 23:04
• This looks awfully elegant to me! I like the way you can eyeball each and every step. Jul 8, 2015 at 1:10

Others may bristle at this "proof," but:

$$7^{31} = 157,775,382,034,845,806,615,042,743 \\ 8^{29} = 154,742,504,910,672,534,362,390,528$$

If all else fails, just calculating the expressions and comparing them will work. This particular problem is only mildly tedious to attack this way if you have pen/paper.

• No need for scare quotes. That's a perfectly valid proof. Jul 7, 2015 at 16:26
• This is in a sense a valid proof, but I can't see how you got the numbers or why I should believe you got them right. This is a proof, but not an easily human-verifiable one. Jul 7, 2015 at 16:38
• Slightly more readable: $7^{31} =101202171144551276401762715267_8$.
– user65203
Jul 7, 2015 at 16:54
• Not for me... similar situation math.stackexchange.com/questions/562538/… Jul 7, 2015 at 17:06
• And for those of us who once memorized a few logarithms: $\log 7\approx0.84510$, and $\log 8\approx0.90309$. By an easy hand computation $31\cdot0.84510=26.1981$, and $28\cdot0.90309=26.18961$. The difference between $26.1981$ and $26.18961$ is too large to be the result of roundoff error. Jul 7, 2015 at 22:48

Here's an idea for a proof that uses some, but hopefully not too much, arithmetic. If you know the power series

$$-\ln(1-x)=x+{1\over2}x^2+{1\over3}x^3+\cdots$$

then you can get started by finagling the desired inequality as follows:

\begin{align} 7^{31}\gt8^{29}&\iff31\ln7\gt29\ln8\\ &\iff31\ln(8-1)\gt29\ln8\\ &\iff31\left(\ln8+\ln\left(1-{1\over8} \right) \right)\gt29\ln8\\ &\iff2\ln8\gt-31\ln\left(1-{1\over8} \right)\\ &\iff6\ln\left(1-{1\over2} \right)\gt-31\ln\left(1-{1\over8} \right)\\ &\iff6\left({1\over2}+{1\over8}+{1\over48}+{1\over64}+\cdots \right)\gt31\left({1\over8}+{1\over128}+{1\over1536}+\cdots \right) \end{align}

The final ingredient is to use the inequality

$${1\over n}x^n+{1\over n+1}x^{n+1}+\cdots\lt{1\over n}\left(x^n+x^{n+1}+\cdots \right)={x^n\over n(1-x)}$$

in truncating the infinite sum on the right. It may take a couple of attempts to find truncations that work.

Added later (after seeing math110's answer): I had quite forgotten my own answer (from two years ago) to the problem of proving $\sqrt7^\sqrt8\gt\sqrt8^\sqrt7$. In it, I showed all the steps necessary to establish

$$-\ln\left(1-{1\over8} \right)\lt{137\over1024}\quad\text{and}\quad6\ln2\gt{1063\over256}$$

So all that remains here is to note that

$$4\cdot1063=4252\gt4247=31\cdot137$$

Whew!

• It's too complicated for me, but thanks for your anwser. Jul 7, 2015 at 18:39
• You could use $$-31\ln\left(\frac78\right) = 31\ln\left(\frac87\right) = 31\left(\frac17 - \frac1{98} + \dots\right)$$, which is an alternating series and thus easily bounded. Jul 8, 2015 at 1:24

The expression

$$7^{31}>8^{29}$$

Is equivalent to $$31\ln(7)>28\ln(8)$$ where $\ln$ denotes the natural logarithm. As $7>e$, this is equivalent to $$\frac{31}{28}>\frac{\ln(8)}{\ln(7)}$$

The above relation can then be easily verified by calculator.

Alternatively, along a similar vein

$$7^{31}=\left(7^{\frac{31}{29}}\right)^{29}$$

As $7^{\frac{31}{29}}\approx8.01>8$ (via my pocket calculator), the inequality follows.

Basically I am still showing this through computation, I'm just trying to make the computations a bit nicer.

• How do you establish your last line? Jul 7, 2015 at 23:46
• Sorry, which line do you mean? Jul 7, 2015 at 23:59
• How do you know $7^{31/29}\approx 8.01$? Jul 8, 2015 at 0:00
• I just used a calculator. I am still proving this by computation, I just tried to make the computations a bit simpler. I'll edit my post to make that clearer. Jul 8, 2015 at 0:03
• John's answer is a much simpler example of using a calculator. Jul 8, 2015 at 0:05

We have to prove $(\frac {7}{8})^{29}>\frac {1}{49}$. Write

$(\frac {7}{8})^{29}=(1-\frac 18)^{29}=[1- (\frac 18)^{29}]+\binom {29} {1}[1-(\frac 18)^{28}]+….+\binom {29} {14}[1-(\frac18)^{14}]$

Just the first term in this sum of positive is already greater than $\frac {1}{49}$. One has $[1- (\frac 18)^{29}]>\frac {1}{49}\iff 8^{29}-1 > \frac {8^{29}}{49}$ which is quite clear.

• Is this the binomial formula? What's happening in this answer? Sep 11, 2017 at 14:35
• @G Tony Jacobs: Do you doubt that $1-\dfrac{1}{49}\gt (\dfrac 18)^{29}\large?$ Sep 11, 2017 at 18:36
• No, I don't. That has nothing to do with my comment. I'm just asking what formula you used here. Sep 11, 2017 at 18:37
• $(a+b)^n=[a^n+b^n]+\binom {n} {1}[a^{n-1}b+ab^{n-1}]+\binom {n} {2}[a^{n-2}b^2+a^2b^{n-2}]….....$ Sep 11, 2017 at 18:47
• Unfortunately, the sum is pretty much all wrong. Apart from the first term, the subsequent ones are missing powers of $(-1/8)$ in the first addend and the signs between the addends should alternate. The terms in the corrected sum are clearly not all positive - in fact, if they were, wouldn't it be strange that $(7/8)^{29}$ is almost equal to $1$ (the first term in the sum) or even much greater than $1$ (second term)? Mar 25, 2021 at 18:57

First, $$7^5=16807$$ and $$2^{14}=16384$$. And we have $$\dfrac{7^{31}}{8^{29}}=\left(\dfrac{7^5}{2^{14}}\right)^6\times\dfrac78$$.

Now, for $$x\ge0$$ and integer $$n\ge0$$, we have $$(1+x)^n\ge1+nx$$, hence

$$\left(\frac{7^5}{2^{14}}\right)^6\ge1+6\left(\frac{7^5}{2^{14}}-1\right)=1+\frac{6\times423}{2^{14}}=1+\frac{3\times423}{2^{13}}=\frac{9461}{8192}$$

Finally,

$$\frac{9461}{8192}\times\frac78=\frac{66227}{65536}>1$$

Therefore,

$$\frac{7^{31}}{8^{29}}>1$$

• It is nice. (+1) Dec 21, 2021 at 11:41

Lemma. $$5^{15} > 2^{32}\cdot7.$$

Proof. The first eight entries in the sixteenth row of Pascal's Triangle are easily computed recursively by hand (or may be read from a table). Using them, the Binomial Theorem gives $$\begin{gather*} \frac{5^{15}}{2^{30}} =\left(1 + \frac14\right)^{15} > \sum_{i=0}^7\frac{\binom{15}i}{4^i} = 1 + \frac{15}4 + \frac{105}{4^2} + \frac{455}{4^3} + \frac{1365}{4^4} + \frac{3003}{4^5} + \frac{5005}{4^6} + \frac{6435}{4^7} \\ = \frac{6435 + 4(5005 + 4(3003 + 4(1365 + 4(455 + 4(105 + 4(15 + 4))))))}{4^7} \\ = \frac{6435+4(5005+4(3003+4(1365+4\cdot1179)))}{4^7} = \frac{6435+4(5005+4(3003+4\cdot6081))}{4^7} \\ = \frac{6435+4(5005+4\cdot27327)}{4^7} = \frac{6435+4\cdot114313}{4^7} = \frac{463687}{4^7} > 4\cdot7, \end{gather*}$$ because $$4^8\cdot7 = 65536\cdot7 = 458752 < 463687.\ \square$$

We have $$7^4 = 2401 > 2400 = 2^5\cdot3\cdot5^2,$$ so $$7^{32} > 2^{40}\cdot3^8\cdot5^{16}.$$

But $$3^8\cdot5 = 6561\cdot5 = 32805 > 32768 = 2^{15},$$ so $$7^{32} > 2^{55}\cdot5^{15}.$$

Now the lemma gives $$7^{32} > 2^{87}\cdot7,$$ whence $$7^{31} > 2^{87} = 8^{29}.\ \square$$

• N.B. The mention of Pascal's Triangle was not intended to suggest (as I've just realised with horror it might) that one has to compute the first fifteen rows before computing the sixteenth! What I had in mind was simply the recursion $$\binom{n}{i+1} = \binom{n}{i}\cdot\frac{n-i}{i+1} \quad (i = 0, 1, 2. \ldots, n - 1).$$ Dec 21, 2021 at 0:51

I tried just approximate the numbers and I think I found a very ugly way to prove the inequality. Are there any mistakes?

We have $7^4>24\cdot10^2$ so $7^{31}=7^3\cdot(7^4)^7>7^3\cdot(24\cdot10^2)^7=2^{35} \cdot 3^7 \cdot 5^{14} \cdot 7^3$. Also $8^{29}=2^{87}$. So one has to prove the inequalities $2^{52}<3^7 \cdot 5^{14} \cdot 7^3=750141\cdot5^{14}<750142\cdot5^{14}$ or $2^{51}<375071\cdot5^{14}<375072\cdot5^{14}$.

This mean we have to prove that $2^{51}<2^5 \cdot 3 \cdot 3907\cdot5^{14}$ or $2^{46}<3\cdot3907\cdot5^{14}<3\cdot3908\cdot5^{14}=3\cdot2^2\cdot977\cdot5^{14}$ or equivalently $2^{44}<3\cdot977\cdot5^{14}$. But $3\cdot977\cdot5^{14}<3\cdot976\cdot5^{14}=3\cdot2^4\cdot61\cdot5^{14}$ so one has to prove that $2^{40}<3\cdot61\cdot5^{14}$.

But $3\cdot61\cdot5^{14}<3\cdot62\cdot5^{14}$ so one has to prove that $2^{40}<3\cdot62\cdot5^{14}$ or $2^{39}<3\cdot31\cdot5^{14}$. But $3\cdot31\cdot5^{14}<3\cdot2^5\cdot5^{14}$ so one has to prove that $2^{34}<3\cdot5^{14}$.

Now, lets approximate some square roots:

The inequality above is equivalent to $2^{17}<\sqrt{3}\cdot5^7$. But $\sqrt{3}>1.7$ so one has to prove that $2^{17}<1.7\cdot5^7$. This is the same as $\sqrt{2}\cdot2^8<\sqrt{1.7\cdot5}\cdot5^3$ or $256\cdot\sqrt{2}<\sqrt{8.5}\cdot125$. This is the same as $(256/125)^2<8.5/2$ or $256^2\cdot2<8.5\cdot125^2$ or $131072<132812.5$.

• This proof doesn't work. You proved that $2^{52}<750142\cdot 5^{14}$, which doesn't imply that $2^{52}<750141\cdot 5^{14}$. Jul 7, 2015 at 23:48