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In this problem, I guess b is larger, but not know how to prove it without going to lengthy calculations. It is highly appreciated if anyone can give me a help.

Which number is larger

$$\begin{align} &\textrm{(a)}\quad 7^{94} &\quad\textrm{(b)}\quad 9^{91} \end{align}$$

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    $\begingroup$ The quick way to do it, especially if it's multiple choice, is comparing the progressions $7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7....$ with $9 \cdot 9 \cdot 9 \cdot 9....$. Intuitively, considering the "growth rate", it is clear that for large $n$ $$\prod_{i=1}^{n-3} 9> \prod_{i=1}^{n} 7$$ $\endgroup$ – MathematicsStudent1122 May 16 '16 at 23:49
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    $\begingroup$ FSK, could you please add something about the fact that you weren't asking for the numerical values, but some quick way? I cannot help but feel bad about the second most upvoted answer. $\endgroup$ – Vincenzo Oliva May 17 '16 at 9:31
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The first is $7^{91}\times 343$. The second is $7^{91}\times(9/7)^{91}$. Since $\frac{9^3}{7^3}\gt 2$, it follows that $(9/7)^{91}$ is much much bigger than $343$.

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  • $\begingroup$ I would not assume that everybody knows that $(9/7)^3>2$. I think most people could not check that without pen and paper. $\endgroup$ – Pakk May 17 '16 at 13:26
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    $\begingroup$ @Pakk: True, I used a fairly tight estimate. But there is a tremendous amount of slack here, as long as we know that a smallish power of $9/7$ is greater than $2$, the inequality follows. If we use $\frac{9^3}{7^3}\gt 2$, then $(9/7)^{91}\gt 2^{30}\gt 1000^3$. $\endgroup$ – André Nicolas May 17 '16 at 16:02
  • $\begingroup$ @Pakk: My answer gives a similar approach using a looser estimate that's trivial to see, if you prefer zero explicit computation. $\endgroup$ – user21820 May 17 '16 at 16:05
  • $\begingroup$ $9^2/7^2=81/49>80/50=1.6 >\sqrt 2$ so $9^4/7^4>2$. $\endgroup$ – DanielWainfleet May 17 '16 at 19:04
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    $\begingroup$ @Pakk I tried to see it as $9^3 > 2 \cdot 7^3$, which is $729 > 2 \cdot 343=686$ $\endgroup$ – Cookie May 17 '16 at 23:02
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$$7^{94} = 7^{10} 49 ^{42} < 7^{10} 54 ^{42} = 7^{10} 8^{14} 9^{63} < 9^{10} 9^{14} 9^{63} = 9^{87} < 9^{91} $$

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    $\begingroup$ Nice exploitation of factorization. $\endgroup$ – André Nicolas May 17 '16 at 15:08
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    $\begingroup$ Alternatively, $7^{94}=49^{47}\lt54^{48}=8^{16}9^{72}\lt9^{88}\lt9^{91}$. $\endgroup$ – Barry Cipra May 17 '16 at 16:18
  • $\begingroup$ Probably the smartest way. $\endgroup$ – Vincenzo Oliva May 17 '16 at 16:22
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$Log(9^{91})=91\cdot Log(9)=86.836068359$

$Log(7^{94})=94\cdot Log(7)=79.4392157613$.

Hence $ 9^{91}$ is bigger.

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    $\begingroup$ This should be the selected answer. $\endgroup$ – Brian Risk May 17 '16 at 1:32
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    $\begingroup$ This doesn't aid the asker in finding a way to obtain solutions for such problems without lengthy calculations (which is what logs boil down to). "Plug this into a calculator thusly" << using simple concepts to give an elegant and concise answer which doesn't need any hard work at all. $\endgroup$ – Nij May 17 '16 at 3:33
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    $\begingroup$ Holy crap, MSE has become embarrassing. $\endgroup$ – Vincenzo Oliva May 17 '16 at 9:14
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    $\begingroup$ @Loffen No insight, either. $\endgroup$ – G. Bach May 17 '16 at 11:41
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    $\begingroup$ This answer should simply be deleted, it's an embarassment for MSE. It currently has 32 upvotes, which tells much about the mathematical level of some MSE users. $\endgroup$ – Alex M. May 17 '16 at 13:05
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André already nailed it, but here's another way. The following inequalities are equivalent:\begin{align}7^{94} &< (7+2)^{94-3} \\ 9^3&<(1+2/7)^{94} \\ 3\log3&<47\log(1+2/7),\end{align} and by the Maclaurin expansion of $\log(1+x)$, the latter follows from \begin{align}3\log3&<94\left(\frac17-\frac1{49}\right) \\ \log3<3&<2\cdot\frac{94}{49}.\end{align}

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    $\begingroup$ I guess the downvoter upvoted guestDiego's answer. $\endgroup$ – Vincenzo Oliva May 17 '16 at 9:27
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    $\begingroup$ This should be the preferred answer to the OPs question. The calculator based answer does not constitute an appropriate answer, despite 16 voters opinions. Well reasoned Vincenzo. $\endgroup$ – Kevin May 27 '16 at 8:39
  • $\begingroup$ So many fans of guestDiego! $\endgroup$ – Vincenzo Oliva Aug 5 '16 at 7:12
  • $\begingroup$ Quite incredible - I stand by my comment to you in May, this is excellent reasoning. Barry's answer is also suitable, but i stick to my vote! All the best, Sir. $\endgroup$ – Kevin Aug 5 '16 at 8:51
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$9^{91} \div 7^{94} = (\frac97)^{94} \div 9^3 > (1+\frac27)^{7 \times 13} \div 3^6 > (1+2)^{13} \div 3^6 = 3^7$ which is way bigger than $1$.

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First notice that $3^9 = 19683 > 16807 = 7^5$ (this can be calculated manually). Thus $9^9 = (3^2)^9 = 3^{18} = (3^9)^2 > (7^5)^2 = 7^{10}$. It follows that $9^{91} > 9^{90} = (9^9)^{10} > (7^{10})^{10} = 7^{100} > 7^{94}$.

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  • $\begingroup$ What... that's not a Suslin tree! :-) $\endgroup$ – Asaf Karagila May 17 '16 at 21:05
  • $\begingroup$ @AsafKaragila Not only is it not a Souslin tree, but the numbers are finite. When was the last time either of us solved a problem involving finite numbers? At least this way it doesn't require AC. $\endgroup$ – Ari Brodsky May 30 '16 at 23:04
  • $\begingroup$ Ha! Well, finite things are always harder than choiceless things, which are inherently harder than choice-y things anyway. :-P $\endgroup$ – Asaf Karagila May 30 '16 at 23:06
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I voted for André's answer, but here's another approach, using a different bit of maths.

Note that $7^{94} = 7^3 \times 7^{91}$. $9^{91} = (7 \times \frac{9}{7})^{91}$, where $\alpha = \frac{9}{7} = 1 + \frac{2}{7} > 1$. So $$ \frac{7^{94}}{9^{91}} = \frac{7^3}{\alpha^{91}}. $$

What do we make of $\frac{7^3}{\alpha^{91}}$? Well, $7^3 = 49 \times 7 = 343$. Using the binomial theorem, and observing that positive ratios always diminish when the numerators (resp. denominators) are decreased (resp. increased), \begin{align*} \alpha^{91} &= \left(1 + \frac{2}{7}\right)^{91} \\ &> 1 + \frac{91}{1!} \times \frac{2}{7} + \frac{91 \times 90}{2!} \times \frac{2^2}{7^2} \\ &\quad= 1 + \frac{182}{7} + \frac{8190 \times 4}{98} \\ &\quad> 1 + 25 + \frac{4 \times 80 \times 100}{100} \\ &\qquad= 1 + 25 + 320 \\ &\qquad= 346 \\ &\qquad> 343. \end{align*} So $\alpha^{91} > 7^3$ and thus $9^{91} > 7^{94}$.

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  • $\begingroup$ Okay, so I just spotted that I made a bit of a meal of the cancellation up top. I also wanted to add that I am positively teeming with news about the binomial theorem. ahem $\endgroup$ – Wooster May 17 '16 at 12:50
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$$\begin{align} \left(9\over7\right)^3={729\over343}\gt2 &\implies\left(9\over7\right)^{15}\gt2^5\gt7\\ &\implies9^{15}\gt7^{16}\\ &\implies9^{90}\gt7^{96}\\ &\implies9^{91}\gt7^{94} \end{align}$$

Another proof, using the general inequality $\ln(1-x)\lt-x$ for $0\lt x\lt1$ and the numerical inequality $7\lt2^3\lt e^3$:

$$\ln(7^{94}/9^{91})=3\ln7+91\ln\left(1-{2\over9}\right)\lt3\cdot3-90\cdot{2\over9}=9-20\lt0$$

A third proof, presented in easily checkable, but almost completely unmotivated form:

$$\begin{align} 2^{47}7^{94} &=98^{47}\\ &\lt100^{47}\\ &=10^3\cdot10^3\cdot10^{88}\\ &\lt2^{10}\cdot2^{10}\cdot10000^{22}\\ &\lt20000^{22}\\ &\lt160^{44}\\ &\lt36(162)^{45}\\ &=2^{47}9^{91} \end{align}$$

And one more proof, this one based on the fact that $2^{10}=1024\gt1000=10^3$, which implies $\log2\gt0.3$ (where "log" here means log base $10$):

$$94\log7=47\log49\lt47(\log100-\log2)\lt47\cdot1.7=79.9$$ whereas

$$91\log9\gt91\log8=273\log2\gt273\cdot0.3=81.9$$

Full disclosure: I used a calculator for $47\cdot1.7=79.9$. But everything else I did by hand.

Added 5/25/15: At another question, proofs are given of the inequality $7^{19}\lt9^{17}$. (See in particular joriki's answer there.) It follows that

$$7^{94}\lt7^{95}\lt9^{85}\lt9^{91}$$

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Here are a couple of equivalent statements: \begin{eqnarray*} 7^{94}&<&9^{91}\\ 7^3&<&\left(\frac97\right)^{91}\\ 343&<&\left(1+\frac27\right)^{91} \end{eqnarray*} We can expand the latter expression using the binomial theorem to get $$\left(1+\frac27\right)^{91}>1+\binom{91}{1}\times\frac27+\binom{91}{2}\times\left(\frac27\right)^2=1+26+\frac{90\times26}{7}>343.$$

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