# Which of the numbers is larger: $7^{94}$ or $9^{91}$?

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}

• 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 $n$ $$\prod_{i=1}^{n-3} 9> \prod_{i=1}^{n} 7$$ May 16 '16 at 23:49
• 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. May 17 '16 at 9:31

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 $343$.

• I would not assume that everybody knows that $(9/7)^3>2$. I think most people could not check that without pen and paper.
– user193810
May 17 '16 at 13:26
• @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 $(9/7)^{91}\gt 2^{30}\gt 1000^3$. May 17 '16 at 16:02
• @Pakk: My answer gives a similar approach using a looser estimate that's trivial to see, if you prefer zero explicit computation. May 17 '16 at 16:05
• $9^2/7^2=81/49>80/50=1.6 >\sqrt 2$ so $9^4/7^4>2$. May 17 '16 at 19:04
• @Pakk I tried to see it as $9^3 > 2 \cdot 7^3$, which is $729 > 2 \cdot 343=686$ May 17 '16 at 23:02

$$7^{94} = 7^{10} 49 ^{42} < 7^{10} 54 ^{42} = 7^{10} 8^{14} 9^{63} < 9^{10} 9^{14} 9^{63} = 9^{87} < 9^{91}$$

• Nice exploitation of factorization. May 17 '16 at 15:08
• Alternatively, $7^{94}=49^{47}\lt54^{48}=8^{16}9^{72}\lt9^{88}\lt9^{91}$. May 17 '16 at 16:18
• Probably the smartest way. May 17 '16 at 16:22

$Log(9^{91})=91\cdot Log(9)=86.836068359$

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

Hence $9^{91}$ is bigger.

• This should be the selected answer. May 17 '16 at 1:32
• 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.
– Nij
May 17 '16 at 3:33
• Holy crap, MSE has become embarrassing. May 17 '16 at 9:14
• @Loffen No insight, either. May 17 '16 at 11:41
• 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. May 17 '16 at 13:05

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}

• I guess the downvoter upvoted guestDiego's answer. May 17 '16 at 9:27
• 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. May 27 '16 at 8:39
• So many fans of guestDiego! Aug 5 '16 at 7:12
• 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. Aug 5 '16 at 8:51

$9^{91} \div 7^{94} = (\frac97)^{94} \div 9^3 > (1+\frac27)^{7 \times 13} \div 3^6 > (1+2)^{13} \div 3^6 = 3^7$ which is way bigger than $1$.

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 $9^{91} > 9^{90} = (9^9)^{10} > (7^{10})^{10} = 7^{100} > 7^{94}$.

• What... that's not a Suslin tree! :-) May 17 '16 at 21:05
• @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. May 30 '16 at 23:04
• Ha! Well, finite things are always harder than choiceless things, which are inherently harder than choice-y things anyway. :-P May 30 '16 at 23:06

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 $9^{91} > 7^{94}$.

• 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 May 17 '16 at 12:50

\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

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

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 $$\left(1+\frac27\right)^{91}>1+\binom{91}{1}\times\frac27+\binom{91}{2}\times\left(\frac27\right)^2=1+26+\frac{90\times26}{7}>343.$$