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This is the one of my tries: $5=2^{\log_{\ 2} 5}$. Then I should prove that: ${\sqrt 7} > {\log_2 5}$.

So, can you help me end this proof or suggest another?

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    $\begingroup$ To prove that $2^{\sqrt 7}$ is larger than $5$, you can estimate $\sqrt 7$ and then estimate $2^{\sqrt 7}$. $\endgroup$ – Pedro Tamaroff Dec 6 '15 at 15:17
  • $\begingroup$ $\sqrt 7=2,645....> 2,5=\frac 52$. And $2^{\frac 52}=[\sqrt 2]^5\approx 5,6568$ $\endgroup$ – Piquito Dec 6 '15 at 15:34
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Hint: $\sqrt{7}>\frac{5}{2}$.

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    $\begingroup$ Lol, my answer is worse than this. To see $\sqrt7>\frac{5}{2}$ just square both sides, it is equivalent to saying $7>\frac{25}{4}=6.25$ which is clear. $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 15:31
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    $\begingroup$ @dREaM $\left(\frac{5}{2}\right)^2=\frac{25}{4}=6.25$. $\endgroup$ – kccu Dec 6 '15 at 15:32
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    $\begingroup$ Oh yeah, thanks for that. $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 15:34
  • $\begingroup$ How did he got ${5/2}$? I know it's connected to ${log_25}$ but how? $\endgroup$ – False Promise Dec 6 '15 at 15:43
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    $\begingroup$ Since $\sqrt{7}>5/2$, it's enough to show $2^{5/2}>5$ which is equivalent to $2^5>5^2$. $\endgroup$ – Wojowu Dec 6 '15 at 15:47
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Note that

$$5\lt2^\sqrt7\iff5^\sqrt7\lt(2^\sqrt7)^\sqrt7=2^7=128$$

But clearly $\sqrt7\lt3$, so $5^\sqrt7\lt5^3=125$.

Added later: Going beyond what the OP requests, here's a proof that $6\lt2^\sqrt7$, in enough detail that everything can be checked, I think, without resorting to a calculator.

Note first that

$$2\lt1+{7\over6}\lt\left(1+{1\over6}\right)^7=\left(7\over6\right)^7\implies 2\cdot6^7\lt7^7$$

and

$$3^5=243\lt256=2^8\implies3^5\cdot3^7\lt2^8\cdot3^7\implies3^{12}\lt2\cdot6^7$$

Putting these together, we have

$$3^{12}\lt7^7$$

From here, using the easy inequality $5/2\lt\sqrt7$ (which can be seen from $5^2=25\lt28=4\cdot7$) and the equality $(\sqrt7+1)(\sqrt7-1)=6$, we have

$$\begin{align} 3^{12}\lt7^7 &\implies3^6\lt7^{7/2}\\ &\implies3^6\lt7^{\sqrt7+1}\\ &\implies3^{(\sqrt7-1)(\sqrt7+1)}\lt7^{\sqrt7+1}\\ &\implies3^{\sqrt7-1}\lt7\\ &\implies3^{\sqrt7+1}\lt9\cdot7=63\lt64=2^6=2^{(\sqrt7-1)(\sqrt7+1)}\\ &\implies3\lt2^{\sqrt7-1}\\ &\implies6\lt2^\sqrt7 \end{align}$$

If there's an appreciably easier proof, I'd be interested to see one; I was a little surprised this one took as many steps as it did.

Added yet later: Here's an appreciably easier proof, using the fact that $7\lt64/9$ implies $\sqrt7\lt8/3$ and the inequality $3^8\lt2^{13}$, which can be verified by noting

$$3^8=81^2\lt90^2=8100\lt8160=8\cdot1020\lt8\cdot1024=2^{13}$$

Putting these together, we have

$$\begin{align} 3^8\lt2^{13} &\implies2^8\cdot3^8\lt2^8\cdot2^{13}\\ &\implies6^8\lt2^{21}\\ &\implies6^{8/3}\lt2^7\\ &\implies6^\sqrt7\lt2^7\\ &\implies6\lt2^\sqrt7 \end{align}$$

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This solution is clearly worse, but it just goes to prove if you try enough stuff something will work.

You want to prove $a<b$, so prove $2^{(a^2)}<2^{(b^2)}$.

In this case we want $2^{(\log_2{5})^2}<2^7$

Of course $(\log_2{5})^2<3(\log_25)$ (since $2^3=8>5)$.

So it suffices to prove $(2^{log_2{5}})^3<2^7$. But of course $125<128$

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  • $\begingroup$ But it seems it is better to just bound $\sqrt{7}$ with something you can work with. $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 15:25
  • $\begingroup$ I think you meant to write $(2^{\log_2(5)})^3<2^7$? (Not greater than...) $\endgroup$ – kccu Dec 6 '15 at 15:31
  • $\begingroup$ Yeah, thanks. Although that is what we wanted to prove. $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 15:33
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Good start. So you need to prove that the square of the 2-log of 5 is less than 7, or equivalently, that the square of the double of the 2-log of 5 is less than 28.

But the square of the double of the 2-log of 5 is actually the square of the 2-log of 25. The latter is less than 5 (because the 2-log of 32 is 5) so its square is less than 25, a fortiori less than 28.

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Hint: use $5^\sqrt{7}<5^3$.

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If you wanna go by without calculator then write$\sqrt{7}=\sqrt{8-1}$ then use binomial theorem for rational powers which i assume you know and then do floor function or GIF function greatest integer function $[x]$ and you are done with your work.

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  • $\begingroup$ How exactly would the multinomial theorem be used here? $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 15:37
  • $\begingroup$ First, I suppose you mean binomial series. Second, the exponent is positive. $\endgroup$ – Wojowu Dec 6 '15 at 15:37
  • $\begingroup$ I edited (rational). $\endgroup$ – Archis Welankar Dec 6 '15 at 15:42
  • $\begingroup$ Eg If we are given find cube root of $7.995$ we generally use binomials or multinomials @dream $\endgroup$ – Archis Welankar Dec 6 '15 at 15:45

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