Can $\sqrt{8}^{\sqrt{7}}$ Ever Be Greater Than $\sqrt{7}^{\sqrt{8}}$?

[anemone]

Prove that $\sqrt{8}^{\sqrt{7}}<\sqrt{7}^{\sqrt{8}}$.

 

[anemone]

Prove that $\sqrt{8}^{\sqrt{7}}<\sqrt{7}^{\sqrt{8}}$.

Subtle Hint:

Spoiler

$7\cdot 31^2<8\cdot k^2$

 

[anemone]

Subtle Hint:

Spoiler

$7\cdot 31^2<8\cdot k^2$

Second subtle hint:

Spoiler

$8^{?}<7^{31}$

 

[anemone]

Prove that $\sqrt{8}^{\sqrt{7}}<\sqrt{7}^{\sqrt{8}}$.

Solution of other:

Spoiler

If we want to prove it with the approach of integer arithmetic, we need to find a rational approximation of \(\displaystyle \sqrt{\frac{7}{8}}\) that is slightly bigger than itself.

We find that $7\cdot 31^2(=6727)<8\cdot 29^2(=6728)$ which implies

$\sqrt{7}\cdot 31<29\cdot \sqrt{8}$

Also, we see that

$8^{29}\approx 1.55\times 10^{26}<7^{31}\approx 1.58\times 10^{26}$

It follows that $(8^{\sqrt{7}})^{31}<(8^{29})^{\sqrt{8}}<(7^{31})^{\sqrt{8}}$

Take the $62^{\text{th}}$ root on both sides the result follows.

 

[Albert1]

Prove that $\sqrt{8}^{\sqrt{7}}<\sqrt{7}^{\sqrt{8}}$.

Spoiler

let :$A=\sqrt{8}^{\sqrt{7}},\,\,B=\sqrt{7}^{\sqrt{8}}$
$A^2=8^\sqrt 7,B^2=7^\sqrt 8$
since: $8^7=2097152<7^8=5764801$
$\therefore A^2<B^2$
so the proof is done

 

[anemone]

Spoiler

let :$A=\sqrt{8}^{\sqrt{7}},\,\,B=\sqrt{7}^{\sqrt{8}}$
$A^2=8^\sqrt 7,B^2=7^\sqrt 8$
since: $8^7=2097152<7^8=5764801$
$\therefore A^2<B^2$
so the proof is done

Thanks for participating, Albert...but...

Spoiler

By following your logic, if we have $4^3(=64)<3^4(=81)$, then does that mean $4^{\sqrt{3}}<3^{\sqrt{4}}$ must hold as well?

You know, the example I cited above doesn't work...

 

[Albert1]

Thanks for participating, Albert...but...

Spoiler

By following your logic, if we have $4^3(=64)<3^4(=81)$, then does that mean $4^{\sqrt{3}}<3^{\sqrt{4}}$ must hold as well?

You know, the example I cited above doesn't work...

Spoiler

let:$f(x)=\sqrt{\dfrac{x+1}{x}}$
$g(x)=\dfrac {log(x+1)}{log(x)}$
if $x\in N ,\,\, and \,\,x>1$
the solution of $f(x)>g(x)$,is $x\geq 7$
if $x=2,3,4,5,6 $ then $f(x)<g(x)$

 

[anemone]

Spoiler

let:$f(x)=\sqrt{\dfrac{x+1}{x}}$
$g(x)=\dfrac {log(x+1)}{log(x)}$
if $x\in N ,\,\, and \,\,x>1$
the solution of $f(x)>g(x)$,is $x\geq 7$
if $x=2,3,4,5,6 $ then $f(x)<g(x)$

Hello again Albert,

Spoiler

I'd like to see the proof for $x\in \Bbb{N}$, we have $f(x)>g(x)$ for $x\ge 7$ thanks.:) That work would complete your solution.

 

[Albert1]

Spoiler

let:$f(x)=\sqrt{\dfrac{x+1}{x}}$
$g(x)=\dfrac {log(x+1)}{log(x)}$
if $x\in N ,\,\, and \,\,x>1$
the solution of $f(x)>g(x)$,is $x\geq 7$
if $x=2,3,4,5,6 $ then $f(x)<g(x)$

Spoiler

the graph of $f(x),\,\, and \,\, g(x)$ both approah 1 as $x\rightarrow \infty$ and both are decreasing
$f(x)$ and $ g(x)$ meet at only one point $6<x<7$
and $f(x)<g(x) ,\,\, when \,\, x<7$