12
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ 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
23
$\begingroup$

Hint: $\sqrt{7}>\frac{5}{2}$.

$\endgroup$
6
  • 3
    $\begingroup$ 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. $\endgroup$
    – Asinomás
    Dec 6 '15 at 15:31
  • 1
    $\begingroup$ @dREaM $\left(\frac{5}{2}\right)^2=\frac{25}{4}=6.25$. $\endgroup$
    – kccu
    Dec 6 '15 at 15:32
  • 1
    $\begingroup$ Oh yeah, thanks for that. $\endgroup$
    – Asinomás
    Dec 6 '15 at 15:34
  • $\begingroup$ How did he got ${5/2}$? I know it's connected to ${log_25}$ but how? $\endgroup$ Dec 6 '15 at 15:43
  • 2
    $\begingroup$ 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
15
$\begingroup$

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

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

$\endgroup$
8
$\begingroup$

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 $125<128$

$\endgroup$
3
  • $\begingroup$ But it seems it is better to just bound $\sqrt{7}$ with something you can work with. $\endgroup$
    – Asinomás
    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. $\endgroup$
    – Asinomás
    Dec 6 '15 at 15:33
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ How exactly would the multinomial theorem be used here? $\endgroup$
    – Asinomás
    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). $\endgroup$ 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 $\endgroup$ Dec 6 '15 at 15:45
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.