# Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$

Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $$2,3,4,5$$ and $$6$$ is $$LCM(2,3,4,5,6)=60$$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sqrt[60]{4^{15}}=\sqrt[60]{2^{30}}\\\sqrt[5]{5}=\sqrt[60]{5^{12}}\\\sqrt[6]{6}=\sqrt[60]{6^{10}}=\sqrt[60]{2^{10}\cdot3^{10}}$$ Now how do we compare $$2^{30},3^{20},4^{15},5^{12}$$ and $$6^{10}$$? I can't come up with another approach.

• Well, looking at the graph of $x^{1/x}$ seems relevant.
– lulu
Commented Jan 22, 2022 at 17:17
• Does this answer your question? Comparing $\pi^e$ and $e^\pi$ without calculating them Commented Jan 23, 2022 at 4:32
• @TobyMak This question is tagged algebra-precalculus, so I don't think the one you linked is a good duplicate target.
– dxiv
Commented Jan 23, 2022 at 8:43

Let$$f(x)=x^{1/x}=e^{\log(x)/x}$$and note that $$f(n)=\sqrt[n]n$$, for each $$n\in\Bbb N$$. You have$$f'(x)=\frac{1-\log(x)}{x^2}e^{\log(x)/x},$$which is greater than $$0$$ on $$[1,e)$$ and smaller than $$0$$ on $$(e,\infty]$$. Therefore $$f$$ is strictly increasing on $$[1,e]$$ and strictly decreasing on $$[e,\infty)$$. So, since $$e<3$$ and since $$3<4<5<6$$,$$\sqrt[3]3>\sqrt[4]4>\sqrt[5]5>\sqrt[6]6.$$Besides, $$\sqrt2=\sqrt[4]4$$. And it is easy to compare $$\sqrt2$$ with $$\sqrt[3]3$$; just use the fact that $$\sqrt2^6=8$$ and that $$\sqrt3^6=9$$.

The following fills-in the remaining step in OP's approach.

Now how do we compare $$2^{30},3^{20},4^{15},5^{12}$$ and $$6^{10}$$?

• $$3^{20} = 9^{10} \gt 8^{10}=2^{30}\,$$ which excludes $$\,\sqrt{2} = \sqrt[4]{4}\,$$ as possible maximums;

• $$3^{20} = 9^{10} \gt 6^{10}$$ which excludes $$\,\sqrt[6]{6}\,$$ as a possible maximum;

• $$3^{20} \gt 3^{18} = 27^6 \gt 25^{6} = 5^{12}$$ which excludes $$\,\sqrt[5]{5}\,$$ as a possible maximum.

• Very elegant and clear solution. Commented Jan 23, 2022 at 18:41

You can do pairwise comparisons. For instance, to compare $$\sqrt[4]4$$ with $$\sqrt[5]5$$, you only need to compute $$4^5$$ and $$5^4$$. (And you can rule out $$\sqrt[6]6$$ easily by comparing it with $$\sqrt[3]3$$.)

$$2^{30}= (2^2)^{15}= 4^{15}$$.

$$2^{30} = (2^3)^{10}=8^{10} < 9^{10} = (3^2)^{10} = 3^{20}$$.

$$6^{10} < 8^{10} = 2^{30}$$.

So we have $$6^{10}<2^{30}=4^{15} < 3^{20}$$.

Just need to figure how $$5^{12}$$ fits in.

$$5^{12} = (5^2)^{6}$$ and $$2^{30}=(2^5)^6$$ and $$5^2 = 25 < 32 = 2^5$$ so $$5^{12}=(5^2)^6 < (2^5)^6 = 2^{30}$$.

So we have $$5^{12}<2^{30}=4^{15} < 3^{20}$$.

We just need to compare $$5^{12}$$ to $$6^{10}$$.

$$5^{12} = (5^6)^2$$ and $$6^{10}= (6^5)^2$$ so we have compare $$5^6$$ to $$6^5$$.

Note that $$6^5 = (5+1)^5 =$$

$$5^5 + 5\times 5^4 + 10\times 5^3 + 10\times 5^2 + 5\times 5 + 1=$$

Now $$5^5$$ and $$5\times 5^4 = 5^5$$ but $$10\times 5^3 = 2\times 5^4 <5\times 5\times 5^5 = 5^5$$ and $$10\times 5^2 < 10\times 5^3 < 5^5$$ and $$5\times 5 + 1 = 26 < 5^5$$ so

$$5^5 + 5\times 5^4 + 10\times 5^3 + 10\times 5^2 + 5\times 5 + 1 < 5\times 5^5 = 5^6$$

So $$6^5 < 5^6$$ and

....

$$6^{10}< 5^{12} < 2^{30} = 5^{15} < 3^{20}$$.

....

Oh, wait... we just needed to find the maximum?

Okay. $$6 < 8$$ so $$6^{10} < 8^{10}=2^{30} = 4^{15}$$ so it isn't $$6^{10}$$.

$$5^2 =25 < 32 < 2^5$$ so $$5^{12} = (5^2)^{6} <(2^5)^6 = 2^{30}$$ so it isn't $$5^{12}$$.

$$8 < 9$$ so $$3^{20} = 9^{10} > 8^{10} = 2^{30}=4^{15}$$ so $$3^{20}$$ is the maximum.

No need to compare $$5^{12}$$ to $$6^{10}$$ (although $$5^{12}$$ is larger)

It is also possible to work with logarithms. We may write for each of these numbers $$\log_2 \sqrt2 \ \ = \ \ \log_2 2^{1/2} \ \ = \ \ \frac12 \ \ , \ \ \log_3 \sqrt[3]3 \ \ = \ \ \frac13 \ \ , \ldots , \ \ \log_6 \sqrt[6]6 \ \ = \ \ \frac{1}{6} \ \ .$$ It will be more helpful to our purpose to express these in terms of a common logarithmic base, say, $$\ \log_2 x \ \ :$$

$$\log_2 \sqrt2 \ \ = \ \ \frac12 \ \ , \ \ \log_2 \sqrt[3]3 \ \ = \ \ \frac{\log_3 \sqrt[3]3}{\log_3 \ 2} \ \ = \ \ \frac{1}{3 \ \log_3 \ 2} \ \ ,$$ $$\log_2 \sqrt[4]4 \ \ = \ \ \frac{1}{4 \ \log_4 \ 2} \ \ , \ \ \log_2 \sqrt[5]5 \ \ = \ \ \frac{1}{5 \ \log_5 \ 2} \ \ , \ \ \log_2 \sqrt[6]6 \ \ = \ \ \frac{1}{6 \ \log_6 \ 2} \ \ .$$

We have one evident equality, $$\ \large{\frac{1}{4 \ \log_4 \ 2} \ = \ \frac{1}{4 \ \log_4 \ [4^{1/2}]} \ = \ \frac{1}{4 \ · \ (1/2)} \ = \ \frac{1}{2} } \ \ .$$

For the rest, we need to examine the consequences of some inequalities. (All of the denominators in the ratios we've found are positive, so there are no undesired complications.)

$$2^3 \ < \ 3^2 \ \ \Rightarrow \ \ 2 \ < \ 3^{2/3} \ \ \Rightarrow \ \ \log_3 2 \ < \ \frac23 \ \ \Rightarrow \ \ \frac12 \ < \ \frac{1}{3 \ \log_3 2} \ \ ;$$

$$2^5 \ > \ 5^2 \ \ \Rightarrow \ \ 2 \ > \ 5^{2/5} \ \ \Rightarrow \ \ \log_5 2 \ > \ \frac25 \ \ \Rightarrow \ \ \frac12 \ > \ \frac{1}{5 \ \log_5 2} \ \ ;$$

$$5^6 \ = \ 15625 \ > \ 6^5 \ = \ 7776 \ \ \Rightarrow \ \ 5 \ > \ 6^{5/6} \ \ \Rightarrow \ \ \log_6 5 \ = \ \frac{\log_6 2}{\log_5 2} \ > \ \frac56$$ $$\Rightarrow \ \ \frac{1}{5 \ \log_5 2} \ > \ \frac{1}{6 \ \log_6 2} \ \ .$$

Since $$\ \log_2 x \$$ is an increasing function of $$\ x \$$ on its domain $$\ x \ > \ 0 \ \ , \$$ the chain of inequalities $$\log_2 \sqrt[3]3 \ \ > \ \ \log_2 \sqrt2 \ = \ \log_2 \sqrt[4]4 \ \ > \ \ \log_2 \sqrt[5]5 \ \ > \ \ \log_2 \sqrt[6]6$$ implies $$\sqrt[3]3 \ \ > \ \ \sqrt2 \ = \ \sqrt[4]4 \ \ > \ \ \sqrt[5]5 \ \ > \ \ \sqrt[6]6 \ \ .$$