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

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    $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 7 '15 at 14:25
  • $\begingroup$ 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? $\endgroup$ – abnry Jul 7 '15 at 14:41
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    $\begingroup$ I think maybe this problem is from my old problem when Christian Blatter solution a step math.stackexchange.com/questions/431461/… $\endgroup$ – math110 Jul 7 '15 at 16:42
  • $\begingroup$ Maybe you could try to use the binomial theorem. Note that $7=8-1$ and $8=7+1$. $\endgroup$ – Surb Jul 7 '15 at 16:52
  • $\begingroup$ 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. $\endgroup$ – lhf Jul 7 '15 at 16:52
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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: $\dfrac{7^{31}}{8^{29}}=\dfrac{7^{2}\cdot7^{29}}{8^{29}}=\dfrac{7^{3}}{8}\Big(\dfrac{7}{8}\Big)^{28}=\dfrac{7^{3}}{8}\Big(\dfrac{7^4}{8^4}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{2401}{4096}\Big)^{7} > \dfrac{7^{3}}{8}\Big(\dfrac{2400}{4100}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{24}{41}\Big)^{7}=\dfrac{7^{3}}{8}\dfrac{24}{41}\Big(\dfrac{24}{41}\Big)^{6}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{24^2}{41^2}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1681}\Big)^{3}>\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1683}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{9 \cdot 64} {9\cdot 187}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{64} {187}\Big)^{3}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{11^3 \cdot 17^3 \cdot 41}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{1331 \cdot 17^3 \cdot 41}>\dfrac{2^{18} \cdot 3 \cdot 7^3}{1332 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{111 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{4551 \cdot 17^3}>\dfrac{2^{16} \cdot 7^3}{4557 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{93 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{1581 \cdot 17^2}>\dfrac{2^{16} \cdot 7}{1582 \cdot 17^2}=\dfrac{2^{15}}{113 \cdot 17^2}=\dfrac{2^{15}}{1921 \cdot 17}>\dfrac{2^{15}}{1924 \cdot 17}=\dfrac{2^{13}}{481 \cdot 17}=\dfrac{8192}{ 8177}>1. \quad\square$

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  • $\begingroup$ How did you get 7^3/8 * (2500/4100)^7? $\endgroup$ – mrJoe Jul 7 '15 at 22:51
  • $\begingroup$ ^^^2400/4100^^^ $\endgroup$ – mrJoe Jul 7 '15 at 22:52
  • $\begingroup$ It's an inequality. $\endgroup$ – Edward ffitch Jul 7 '15 at 22:58
  • $\begingroup$ $93\cdot 17^3\equiv 1 \pmod 2$ but $1582\cdot 17^2\equiv 0\pmod 2$. $\endgroup$ – Jaakko Seppälä Jul 7 '15 at 23:04
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    $\begingroup$ This looks awfully elegant to me! I like the way you can eyeball each and every step. $\endgroup$ – Barry Cipra Jul 8 '15 at 1:10
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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.

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    $\begingroup$ No need for scare quotes. That's a perfectly valid proof. $\endgroup$ – Eric M. Schmidt Jul 7 '15 at 16:26
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    $\begingroup$ Slightly more readable: $7^{31} =101202171144551276401762715267_8$. $\endgroup$ – Yves Daoust Jul 7 '15 at 16:54
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    $\begingroup$ Not for me... similar situation math.stackexchange.com/questions/562538/… $\endgroup$ – JP McCarthy Jul 7 '15 at 17:06
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    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jul 7 '15 at 22:48
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    $\begingroup$ @JpMcCarthy: It clearly is a valid proof, and it’s easily (if tediously) checked; claiming otherwise is simply denying the facts. $\endgroup$ – Brian M. Scott Jul 7 '15 at 22:56
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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!

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  • $\begingroup$ It's too complicated for me, but thanks for your anwser. $\endgroup$ – mrJoe Jul 7 '15 at 18:39
  • $\begingroup$ 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. $\endgroup$ – NovaDenizen Jul 8 '15 at 1:24
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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.

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  • $\begingroup$ How do you establish your last line? $\endgroup$ – JP McCarthy Jul 7 '15 at 23:46
  • $\begingroup$ Sorry, which line do you mean? $\endgroup$ – Ruvi Lecamwasam Jul 7 '15 at 23:59
  • $\begingroup$ How do you know $7^{31/29}\approx 8.01$? $\endgroup$ – JP McCarthy Jul 8 '15 at 0:00
  • $\begingroup$ 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. $\endgroup$ – Ruvi Lecamwasam Jul 8 '15 at 0:03
  • $\begingroup$ John's answer is a much simpler example of using a calculator. $\endgroup$ – JP McCarthy Jul 8 '15 at 0:05
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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.

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  • $\begingroup$ Is this the binomial formula? What's happening in this answer? $\endgroup$ – G Tony Jacobs Sep 11 '17 at 14:35
  • $\begingroup$ @G Tony Jacobs: Do you doubt that $1-\dfrac{1}{49}\gt (\dfrac 18)^{29}\large?$ $\endgroup$ – Piquito Sep 11 '17 at 18:36
  • $\begingroup$ No, I don't. That has nothing to do with my comment. I'm just asking what formula you used here. $\endgroup$ – G Tony Jacobs Sep 11 '17 at 18:37
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    $\begingroup$ $(a+b)^n=[a^n+b^n]+\binom {n} {1}[a^{n-1}b+ab^{n-1}]+\binom {n} {2}[a^{n-2}b^2+a^2b^{n-2}]….....$ $\endgroup$ – Piquito Sep 11 '17 at 18:47
  • $\begingroup$ Oooooooh! I see it now. I'm not used to seeing the terms grouped like that; thank you for clarifying. $\endgroup$ – G Tony Jacobs Sep 11 '17 at 18:51
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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 $131072<132812.5$.

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  • $\begingroup$ This proof doesn't work. You proved that $2^{52}<750142\cdot 5^{14}$, which doesn't imply that $2^{52}<750141\cdot 5^{14}$. $\endgroup$ – Batominovski Jul 7 '15 at 23:48

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