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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.

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4  
Hint: try $\log_5 7 < 1.4$. –  Secret Math Jun 18 '13 at 4:59
    
That's the other ways around, since $x\longmapsto 5^x$ is increasing: $\log_57<\sqrt{2}\iff 7<5^\sqrt{2}$. –  1015 Jun 18 '13 at 5:06
    
@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 5:16
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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 5:21
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@A.D Show that $7^5<5^7$ –  Thomas Andrews Jun 18 '13 at 5:21

4 Answers 4

up vote 54 down vote accepted

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.

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I like this better than my answer :) nice job –  A.E Jun 18 '13 at 5:25
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Is there any trick to start or we just go on –  iostream007 Jun 18 '13 at 5:42
    
This will hit the nice answer badge soon, and it will be deserved. –  1015 Jun 18 '13 at 5:44
    
This is a good answer. –  Gastón Burrull Jun 18 '13 at 5:44
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@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 5:52

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

  • $7 < 5^{1.4} <5^{\sqrt{2}}$
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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.

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GREAT ANSWER, howd u come up with the function?? –  Ramanujan Sep 13 at 5:31

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

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