5
$\begingroup$

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.

$\endgroup$
4
  • 4
    $\begingroup$ Well, looking at the graph of $x^{1/x}$ seems relevant. $\endgroup$
    – lulu
    Commented Jan 22, 2022 at 17:17
  • $\begingroup$ Does this answer your question? Comparing $\pi^e$ and $e^\pi$ without calculating them $\endgroup$
    – Toby Mak
    Commented Jan 23, 2022 at 4:32
  • $\begingroup$ Please search your question on Approach0 before asking. $\endgroup$
    – Toby Mak
    Commented Jan 23, 2022 at 4:32
  • 2
    $\begingroup$ @TobyMak This question is tagged algebra-precalculus, so I don't think the one you linked is a good duplicate target. $\endgroup$
    – dxiv
    Commented Jan 23, 2022 at 8:43

5 Answers 5

6
$\begingroup$

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 $\sqrt3^6=9$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Very elegant and clear solution. $\endgroup$
    – user376343
    Commented Jan 23, 2022 at 18:41
2
$\begingroup$

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 $\sqrt[3]3$.)

$\endgroup$
0
$\begingroup$

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

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

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

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

....

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

Final answer.

....

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)

My final final answer.

$\endgroup$
0
$\begingroup$

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 $$ \sqrt[3]3 \ \ > \ \ \sqrt2 \ = \ \sqrt[4]4 \ \ > \ \ \sqrt[5]5 \ \ > \ \ \sqrt[6]6 \ \ . $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .