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

Question

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 $\frac{8^{31}}{8^2}$.

Answer 1

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

Answer 2

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.

Answer 3

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!

Answer 4

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.

Answer 5

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.

Answer 6

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,

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

Answer 7

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 $7^{31} > 2^{87} = 8^{29}.\ \square$

Answer 8

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 $131072<132812.5$.