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

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.

• Hint: try $\log_5 7 < 1.4$. Commented Jun 18, 2013 at 4:59
• That's the other ways around, since $x\longmapsto 5^x$ is increasing: $\log_57<\sqrt{2}\iff 7<5^\sqrt{2}$. Commented Jun 18, 2013 at 5:06
• 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 $x=e$. Commented Jun 18, 2013 at 5:21
• @A.D Show that $7^5<5^7$ Commented Jun 18, 2013 at 5:21
• @AndréNicolas You don't just need a maximum at $e$, you need to know it is decreasing for $x>e$. Commented Jun 18, 2013 at 5:22

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.

• I like this better than my answer :) nice job
– A.E
Commented Jun 18, 2013 at 5:25
• Is there any trick to start or we just go on Commented Jun 18, 2013 at 5:42
• This will hit the nice answer badge soon, and it will be deserved. Commented Jun 18, 2013 at 5:44
• This is a good answer. Commented Jun 18, 2013 at 5:44
• @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. Commented Jun 18, 2013 at 5:52

$f(x)=x^\frac1x$ 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 $$5^\sqrt2>5^{1.4}>7$$

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

• $7 < 5^{1.4} <5^{\sqrt{2}}$

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.

• GREAT ANSWER, howd u come up with the function?? Commented Sep 13, 2014 at 5:31

# 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
Commented Jul 3, 2021 at 2:32
• @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 $\lfloor{\mathtt{NUM}* 10^{k}}\rfloor > \lfloor{\mathtt{num} * 10^{k}}\rfloor$ Commented Jul 3, 2021 at 3:20
• 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
Commented Jul 3, 2021 at 3:39