Which is bigger, ${\sqrt2}^9$ or $9^{\sqrt2}$? (no calculator)

Question

I was thinking about the classic question, "Which is bigger, $e^\pi$ or $\pi^e$?" (no calculator), and I tried to create a question that is similar, but resistant to the usual methods used for the classic question. Here is what I came up with:

Which is bigger, ${\sqrt2}^9$ or $9^{\sqrt2}$ ?

The methods used for the classic question take advantage of the special properties of $e$ (for example, the maximum value of $y=\frac{\log x}{x}$ occurs at $x=e$). But in my question there is no $e$, so it seems that some other method must be used.

I can imagine a solution using Maclaurin series, with a lot of tedious pen-and-paper calculation. But I am mostly interested in solutions that require a minimum of calculation.

(According to my calculcator, $\sqrt2^9\approx 22.63$ and $9^\sqrt2\approx 22.36$.)

Answer 1

I think here is an alternate solution nobody has mentioned $$9^{\sqrt{2}} = (3^2)^{\sqrt{2}} = 3^{2\sqrt{2}},$$ $$\sqrt{2}^9 = (\sqrt{2})^{3 \cdot 3} = (2\sqrt{2})^3.$$ Since $e < 2.8 < 2\sqrt{2} < 3$, we have $\sqrt{2}^9 > 9^{\sqrt{2}}$.

Answer 2

We have that

$${\sqrt2}^9 > 9^{\sqrt2} \iff 2^{9\sqrt 2} >81^2$$

and using that $\sqrt 2>1.414$ and that $2^x>1+x\log 2$ with $\log 2>0.69$

$$2^{9\sqrt 2}>2^{12.7}=2^{12}\cdot 2^{0.5}\cdot 2^{0.2}>4096\cdot1.41\cdot1.138>6572>81^2=6561$$

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We can simplify a little bit the last calculation showing at first that $1.41\cdot 1.138>1.604$ and then $4096\cdot1.41\cdot1.138>4096\cdot 1.604>6569$.

Answer 3

An equivalent formulation of the question is: Which is bigger, $\frac{9\sqrt{2}}{8}$ or $\log_2{3}$? I shall use the approach of finding a rational number that's between the two values.

Part 1: $\frac{9\sqrt{2}}{8} > \frac{27}{17}$

$$578 > 576$$ $$2 > \frac{576}{289}$$ $$\sqrt{2} > \sqrt{\frac{576}{289}} = \frac{24}{17}$$ $$\frac{\sqrt{2}}{8} > \frac{3}{17}$$ $$\frac{9\sqrt{2}}{8} > \frac{27}{17}$$

Part 2: $\log_2 3 < \frac{27}{17}$

This inequality is equivalent to $3^{17} < 2^{27}$. It's a bit tedious to work out by hand, but you can show that

$$129\,140\,163 < 134\,217\,728$$

Part 3: Putting it together

By transitivity, since $\frac{9\sqrt{2}}{8} > \frac{27}{17}$ and $\frac{27}{17} > \log_2 3$

$$\frac{9\sqrt{2}}{8} > \log_2 3$$ $$\frac{9\sqrt{2}}{8} > \frac{\ln 3}{\ln 2}$$ $$9\sqrt{2} \ln 2 > 8 \ln 3$$ $$\frac{9}{2} \ln 2 > 2\sqrt{2} \ln 3$$ $$2^{9/2} > 3^{2\sqrt{2}}$$ $$(2^{1/2})^9 > (3^2)^\sqrt{2}$$ $$\sqrt{2}^9 > 9^\sqrt{2}$$

Answer 4

Since $12^2\cdot 2 = 288 < 289 = 17^2$, we have $12\sqrt{2} < 17$. So $9^{12\sqrt{2}} < 9^{17}$.

$9^{17} < \sqrt{2}^{9\times 12}$? Or $3^{17} < 2^{27}$? Or $3^{18} < 2^{26}\cdot 6$? Or $3^9 < 2^{13}\sqrt{6}$?

Or $3^9 < 2^{13}(2 + 4/9) < 2^{13} \sqrt{6}$? ($2^{13} = 8192$, $3^9 = 19683$.) Yes.

Thus, $\sqrt{2}^9 > 9^{\sqrt{2}}$.

Answer 5

Following the idea of this video, we first observe that $2<\frac{2k+1}{k}$ for all positive integers k. We seek the smallest value of k (greater than $4$, which is "trivial") such that both the numerator and the denominator are perfect squares. It turns out that $k=144$, so $2<\frac{289}{144} \implies \sqrt{2}<\frac{17}{12} \implies 9^\sqrt{2} < 9^\frac{17}{12}$.

To conclude, we want to prove that $9^\frac{17}{12} < \sqrt{2}^9 \iff 9^{17} < \sqrt{2}^{108} \iff 9^{17} < 2^{54} \iff 3^{34} < 2^{54} \iff 3^{17} < 2^{27}$.

I did not come up with a method to prove this inequality, but the OP accepts it. Therefore, the inequality holds by transitivity: $$9^\sqrt{2} < 9^\frac{17}{12} < \sqrt{2}^9$$

EDIT #$1$: The best I could do was to compare powers of $2$ and $3$ and observed that $3^5 = 243 < 256 = 2^8$, which implies $3^{15}< 2^{24}$. However, we cannot conclude $3^{17} < 2^{27}$ from here, since $3^2 = 9 > 8 = 2^3$. I think this is strictly related to @TonyK's answer.

EDIT #$2$: Apparently, the "Dan" whose answer I was referring to is not the same "Dan" who posted the question. Kind of hilarious.

Answer 6

The challenge is to avoid calculators.

The original question is equivalent to asking whether $~\displaystyle 2^{[~9\sqrt{2}~]/2} > 81.$

Assuming that it is also known that $~\log_{10} 2 \approx 0.301, ~\log_{10} 3 \approx 0.477,~$ you can take the logarithm, base 10, of both sides.

Note:
If interested, see step 2 of this answer for a somewhat convoluted approximation/derivation of the above base $~10~$ logarithms.

So, the question is equivalent to asking whether

$$(9/2) \times \sqrt{2} \times 0.301 > 4 \times 0.477.$$

This is equivalent to asking whether

$$2 = \sqrt{2}^2 > \left[ ~\frac{4 \times 0.477}{(9/2) \times 0.301} ~\right]^2.$$

This is equivalent to asking whether

$$2 > \left[ ~\frac{8 \times 0.477}{9 \times 0.301} ~\right]^2 \tag1.$$

The question in (1) above can be simplified by noting that

$$\frac{0.477}{0.301} < \frac{0.477}{0.3} = 1.59.$$

Note:
I am assuming that the over-estimation of $~\dfrac{0.477}{0.301}~$ by $~1.59~$ satisfies any potential $~4$-th decimal place rounding errors, in the assumption that $~\log_{10} 2 \approx 0.301, ~\log_{10} 3 \approx 0.477.$ In other words, I am assuming that $~2^{1.59} > 3.$

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So, the entire problem has been reduced to proving that

$$\frac{64}{81} \times (1.59)^2 < 2 = \frac{162}{81},$$

which is equivalent to proving that

$$64 \times (1.59)^2 < 162. \tag2 $$

Write $~(1.59)^2~$ as

$$(1.60 - .01)^2 = 2.56 - (2 \times .01 \times 1.6) + (0.0001)$$

$$= 2.5600 - 0.0320 + 0.0001 = 2.5281.$$

Then, consider that

$$\frac{162}{64} = \frac{81}{32} = \frac{80}{32} + \frac{1}{32} > \left[ ~\frac{8}{32} \times 10 ~\right] + \frac{1}{33.\overline{3}} = 2.50 + .03 = 2.53.$$

So, you have that

$$\frac{162}{64} > 2.53 > 1.59^2.$$

Therefore, the inequality in (2) above is established, and the problem is solved.

Answer 7

Consider whether the ratio of the two terms is larger or smaller than $1$:

$$\sqrt2 ^{9} /9^ {\sqrt2} = \exp(4.5 \ln2 - 2\sqrt2 \ln3)$$

To test whether the exponential term is positive or negative rewrite it as follows:

$$4.5 \ln2 - 2\sqrt2 \ln3 = (4.5 - 2\sqrt2 \ln3/\ln2)\ln2$$

At this point we are fortunate that $\ln(3)/\ln(2)$ can be accurately approximated as a ratio of two natural numbers, namely as $19/12$. The relative error is just $0.001$. Using this approximation we see that the factor above is proportional to

$$27/19 - \sqrt 2$$

We can easily establish that this quantity is positive ($1.421$ versus $1.414$) with a relative error of $0.005$, i.e. a factor $5$ larger than that made in the approximation. Therefore we can safely conclude that the first expression, $\sqrt2 ^{9}$ is the largest.

Answer 8

The problem reduces to race of powers of $2$ and $3$: $$\sqrt2^9>9^\sqrt2\iff2^9>9^\sqrt8$$ Since $\frac{17}6>\sqrt8\iff 289>288$, it is enough to show that $$2^{27}>3^{17}$$ The last inequality can be verified by hand.

Answer 9

We have that

$$\sqrt 2 >1.4142 \implies \sqrt 2\cdot 9 >1.4142\cdot (10-1)=14.142-1.4142>12.72=1.59\cdot 8$$

therefore

$${\sqrt2}^9 > 9^{\sqrt2} \iff 2^{9\sqrt 2} >2^{12.72}>3^8$$

then it suffices to check

$$2^{1.59}>3 \iff \frac{\log 3}{\log 2}<1.59$$

which is true since $$\frac{\log 3}{\log 2}<\frac{27}{17} \iff 3^{17}<2^{27}$$

Answer 10

Taking logarithms of both the given numbers to the base $10$, it is clear that the question is equivalent to comparing $\frac{\log_{10}2}{2\sqrt2}$ and $\frac{2\log_{10}3}9$. Using the approximations $\log_{10}\approx0.301$, $\log_{10}3\approx0.477$ and $\sqrt2\approx1.414$, the problem becomes

$$\text{Compare }\frac{0.301}{2.828}\text{ and }\frac{0.954}{9}$$ $$\equiv\text{Compare }0.301\times9\text{ and }2.828\times0.954$$ $$\equiv\text{Compare }2.709\text{ and }2.697912$$

Hence, $\boxed{\sqrt2^9>9^\sqrt2}$.