Which number is greater, $11^{11}$ or $9^{12}$?

Question

Which number is greater than $11^{11}$ or $9^{12}$?

My work so far:

$11^{11}=285311670611>9^{12}=282429536481$.

But to verify the validity of equality should be in the range of easily verifiable calculations.

Answer 1

$$ \frac{11^{11}}{9^{11}} =\left(\frac{2}{9}+1\right)^{11} = \sum_{k=0}^{11} \binom{11}{k} \left(\frac{2}{9}\right)^k > \sum_{k=0}^5 \binom{11}{k} \left(\frac{2}{9}\right)^k = \frac{177665}{19683} > 9 $$

Answer 2

This is a variant on Servaes's answer. Note that $3^5=243=2\cdot11^2+1$. Using a binomial expansion and some extremely crude upper bounds, we find

$$\begin{align} 3\cdot9^{12} &=3^{25}\\ &=(2\cdot11^2+1)^5\\ &=32\cdot11^{10}+80\cdot11^8+80\cdot11^6+40\cdot11^4+10\cdot11^2+1\\ &\lt32\cdot11^{10}+80\cdot11^8+11^8+11^8+11^8+11^8\\ &\lt32\cdot11^{10}+121\cdot11^8\\ &=32\cdot11^{10}+11^{10}\\ &=3\cdot11^{11} \end{align}$$

and thus $9^{12}\lt11^{11}$.

Answer 3

The goal is to prove that $(1 + 2/9)^{11} > 9$. As the left-hand side is approximately $9.091843$ this will be a bit tricky.

The big idea in this solution is to try to exploit the fact that $(11/9)^2 = 121/81$ is just under $3/2$, since $3/2$ will be simple to work with.

Start with the inequality $3^2 \times 29 > 2^8$, i. e. $261 > 256$. Multiply throuhg on both sides by $3^4 2^3$ to get $3^6 \times 232 > 81 \times 2^{11}$.

Now, we can rewrite this as

$$ 3^{11} \times {232 \over 243} > 81 \times 2^{11}. $$

We then have $(232/243) = 1-11/243 < (1-1/243)^{11}$ (by Bernoulli's inequality, as pointed out by roby5) and so it follows that

$$ 3^{11} \times (1-1/243)^{11} > 81 \times 2^{11}. $$

At this point most of the work is done. Divide both sides by $2^{11}$ to get

$$ 1.5^{11} \times (1-1/243)^{11} > 81 $$

and multiply both sides by $81^{11}$ to get

$$ 121.5^{11} \times (1-1/243)^{11} > 81^{12} $$

But since both factors on the left-hand side are eleventh powers, we can rewrite this as

$$ 121^{11} > 81^{12} $$

and taking square roots of both sides gives the desired result.

Answer 4

More generally, you are looking to prove: $x^x > (x-2)^{x+1}$. This can be done by taking log both sides, and its easier. Consider $f(x) = x\ln x - (x+1)\ln (x-2)$ on $(11, \infty)$, and taking log we have: $f'(x) = \ln x + 1 - \ln(x-2) - \dfrac{x+1}{x-2}$. We have $f''(x) = \dfrac{1}{x} - \dfrac{1}{x-2}+ \dfrac{3}{(x-2)^2}= \dfrac{(x-2)^2-x(x-2) + 3}{x(x-2)^2}= \dfrac{7-2x}{x(x-2)^2} < 0\implies f'(x) > f'(\infty) = 0 \implies f(x) > f(11) $

Answer 5

Just for the heck of it, here's another approach using the fact that $$9^3=6\times11^2+3\qquad\text{ and }\qquad 6^4<11^3-3\times11.$$ It is by no means the slickest way; \begin{eqnarray*} 9^{12}&=& (6\times11^2+3)^4=3^4\times(2\times11^2+1)^4\\ &=& 3^4\times(2^4\times11^8+4\times2^3\times11^6+6\times2^2\times11^4+4\times2\times11^2+1)\\ &=&6^4\times(11^8+2\times11^6+\tfrac{3}{2}\times11^4+\tfrac{1}{2}\times11^2+\tfrac{1}{16})\\ &<&(11^3-3\times11)\times(11^8+2\times11^6+\tfrac{3}{2}\times11^4+\tfrac{1}{2}\times11^2+\tfrac{1}{16})\\ &=&11^{11}+2\times11^9+\tfrac{3}{2}\times11^7+\tfrac{1}{2}\times11^5+\tfrac{1}{16}\times11^3\\ &\ &\quad\ \ \ -3\times11^9-\ 6\times11^7-\tfrac{9}{2}\times11^5-\ \tfrac{3}{2}\times11^3-\tfrac{3}{2}\times11 \end{eqnarray*} The alignment of the last expression shows that it is smaller than $11^{11}$, as desired.