Prove that $\log _5 7 < \sqrt 2.$

Question

Prove that $\log _5 7 < \sqrt 2.$

Trial : Here $\log _5 7 < \sqrt 2 \implies 5^\sqrt 2 <7.$ But I don't know how to prove this. Please help.

That's the other ways around, since $x\longmapsto 5^x$ is increasing: $\log_57<\sqrt{2}\iff 7<5^\sqrt{2}$. — Julien, Jun 18 '13 at 05:06
@SecretMath: So $\log_5 7<7/5 \implies 5\log_5 7<7 $. Then how I show $5\log_5 7<7 $ — A.D, Jun 18 '13 at 05:16
You need to that $5^7\gt 7^5$. So want $7\ln 5\gt 5\ln 7$, i.e. $\frac{\ln 5}{5}\gt \frac{\ln 7}{7}$. Show (calculus) that $\frac{\ln x}{x}$ reaches a max at $x=e$. — André Nicolas, Jun 18 '13 at 05:21
@AndréNicolas You don't just need a maximum at $e$, you need to know it is decreasing for $x>e$. — Thomas Andrews, Jun 18 '13 at 05:22

score 61 · Accepted Answer · answered Jun 18 '13 at 05:21

61

Observe that: $$ \begin{align*} \log_5 7 &= \dfrac{3}{3}\log_5 7 \\ &= \dfrac{1}{3}\log_5 7^3 \\ &= \dfrac{1}{3}\log_5 343 \\ &< \dfrac{1}{3}\log_5 625\\ &= \dfrac{1}{3}\log_5 5^4\\ &= \dfrac{1}{3}(4)\\ &= \sqrt{\dfrac{16}{9}}\\ &< \sqrt{\dfrac{18}{9}}\\ &= \sqrt{2}\\ \end{align*} $$ as desired.

answered Jun 18 '13 at 05:21

Adriano

41,969

1

I like this better than my answer :) nice job – A.E Jun 18 '13 at 05:25
1

Is there any trick to start or we just go on – iostream007 Jun 18 '13 at 05:42
This will hit the nice answer badge soon, and it will be deserved. – Julien Jun 18 '13 at 05:44
This is a good answer. – Gaston Burrull Jun 18 '13 at 05:44
4

@iostream007 It was a bit of trial and error. First I tried doing: $$ \log_5 7 = \dfrac{1}{2} \log_5 49 < \dfrac{1}{2} \log_5 125 = \dfrac{3}{2} $$ But $3/2 > \sqrt{2}$, so I needed to refined the upper bound. Using $\dfrac{1}{3} \log_5 7^3$ did the trick. – Adriano Jun 18 '13 at 05:52
@Adriano okay I got it now – iostream007 Jun 18 '13 at 06:58

score 5 · Answer 2 · answered Jun 18 '13 at 09:30

$f(x)=x^\frac1x$ is a function defined on $(0,\infty)$ its log is $$G(x)=\log f(x)=\frac{\log x}x$$ $$G'(x)=\frac{1-\log x}{x^2}$$ Therefore $G(x)=\log f(x)$ strictly decreases for $x>e$, but logarithm is monotone on $(0,\infty)$ so that $f(x)$ is strictly decreasing for $x>e$

This gives us $$5^{\frac15}>7^{\frac17}$$ implying (by taking 7th power) that $$5^\sqrt2>5^{1.4}>7$$

score 4 · Answer 3 · answered Jun 18 '13 at 05:23

Want to prove that

$\log_5{7} = \frac{\lg{7}}{\lg{5}} < \sqrt{2}$

Equivalently we can show that

$\lg{7} < \lg{5}\times\sqrt{2}$
$7 < 5^{\sqrt{2}}$

where $\lg$ is the base 2 logarithm. Notice that

$5\times5^{\frac{2}{5}}= 5^{1.4} <5^{\sqrt{2}}$

So can we show that $\frac{7}{5} < 5^{\frac{2}{5}}$? Sure, since $7<8=32768^{\frac{1}{5}}<78125^{\frac{1}{5}}$. Hence

$7 < 5^{1.4} <5^{\sqrt{2}}$

Gaston Burrull · Answer 4 · 2013-06-18T05:46:38.887

3

Define $f(x)=5^{x/5}-x$.

If $x>5$ is obvious that $f'(x)=5^{(-1+x/5)}\ln (5)-1>0.$

Since $f(5)=0$, we have $f(7)>0$.

Since $\frac{7}{5}=\sqrt{\frac{49}{25}}<\sqrt{2}$ we are done.

edited Jun 18 '13 at 05:46

answered Jun 18 '13 at 05:35

Gaston Burrull

5,679

GREAT ANSWER, howd u come up with the function?? – Shobhit Sep 13 '14 at 05:31

score -1 · Answer 5 · answered Jul 03 '21 at 02:25

Prove that $\log _5 7 < \sqrt 2$

Lemma

Mathematicians use the word "lemma" to refer to a fact about math which helps us prove other facts about math.

The following is a useful lemma:

Let $x$ and $y$ be any two positive decimal numbers.

For example,

$x$ could be $\pi \approx 3.141592653589793238462643383282 \cdots$

$x$ could be $\frac{4}{3}$

$x$ could be the number $9$

$x$ could be $0.407129911$

If
there exists a positive whole number $k$ such that $\lfloor{x * 10^{k}}\rfloor > \lfloor{y * 10^{k}}\rfloor$
then
$x > y$

If you are unfamiliar with the notation $\lfloor{x}\rfloor$, it means "delete everything to the right of the decimal point."

For example,

$\lfloor{3.1459}\rfloor = 3$

$\lfloor{45.28}\rfloor = 45$

$\lfloor{1.5}\rfloor = 1$

In other words, if you multiply both decimal numbers by ten a few times, and then throw away everything to the right of the decimal point, then you can sometimes tell which number is bigger.

$ \begin{align} log_5(7) & \approx 1.209061 \dots \\ \sqrt{2} & \approx 1.4142136 \dots \\ \end{align} $

$\begin{align} \lfloor 10*\sqrt{2} \rfloor & = 14 \\ \lfloor 10*\log_5(7)\rfloor & = 12 \\ \end{align} $

$14 = \lfloor 10*\sqrt{2} \rfloor > \lfloor 10*\log_5(7) \rfloor = 12$

Therefore, $\sqrt{2} > log_5{7}$

$log_5(7)≈1.209061$... $\sqrt{2}≈1.4142136$... 1) How do you get those without a calculator? 2) If you know those values, you don't need the lemma to realize the first one is smaller. 3) You cannot take the $\lfloor . \rfloor$ greatest integer part on both sides of an inequality to establish it. — dxiv, Jul 03 '21 at 02:32
@dxiv Your point number (3) is incorrect. You wrote, "You cannot take the $⌊.⌋$ greatest integer part on both sides of an inequality to establish it." You can establish inequalities using the floor-function in way I did in my answer. For any two positive real numbers $\mathtt{NUM}$ and $\mathtt{num}$, $\mathtt{NUM} > \mathtt{num}$ if and only if $\exists k \in \mathbb{N}$ such that $\lfloor{\mathtt{NUM}* 10^{k}}\rfloor > \lfloor{\mathtt{num} * 10^{k}}\rfloor$ — Toothpick Anemone, Jul 03 '21 at 03:20
Point #3 means $\lfloor a \rfloor \le \lfloor b \rfloor ;;;\not!!!!!\iff a \le b;$ which is correct. What you added in the comment is also correct, but the posted answer does not make that point, nor does it explain why you chose to multiply by $10$ instead of $1$ or $1000$. Not to mention that there is no math justification why those decimals are supposed to be correct to begin with. — dxiv, Jul 03 '21 at 03:39

Prove that $\log _5 7 < \sqrt 2.$

5 Answers5

Prove that $\log _5 7 < \sqrt 2$

Lemma

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