Is $10^a\lt 127$ True for the Real Root of $x^3-3x-3=0$?

[anemone]

Given $a$ is the real root for $x^3-3x-3=0$, prove that $10^a\lt 127$.

 

[Albert1]

Given $a$ is the real root for $x^3-3x-3=0$, prove that $10^a\lt 127$.

my solution:

Spoiler

let $f(x)=x^3-3x-3$
$f(2.10351)<0,f(2.10381)>0$
there is a real solution "a" satisfying $2.10351<a<2.10381$
but $log 127\approx 2.10381$
$\therefore 10^a<127$

 

[Greg]

Given $a$ is the real root for $x^3-3x-3=0$, prove that $10^a\lt 127$.

Spoiler

\(\displaystyle a=2\cosh\left(\dfrac13\cosh^{-1}\left(\dfrac32\right)\right)=\left(\dfrac{3+\sqrt5}{2}\right)^{1/3}+\left(\dfrac{3-\sqrt5}{2}\right)^{1/3}<\log(127)\)

...but I don't have proof. :(

 

[anemone]

I made a silly blunder...I thought I solved this problem elegantly and so I posted this problem as a challenge...and now that the time has come for me to post my solution, I realize my solution doesn't work. (Angry)

I have to admit this and I will continue to work on this problem and hopefully, I can crack using a different method than Albert's (thanks Albert for participating) and perhaps work on the idea provided by greg1313.

 

[MarkFL]

My solution:

Spoiler

It will suffice to show that:

\(\displaystyle a<\log(127)\)

If we define:

\(\displaystyle f(x)=x^3-3x-3\)

A quick sign check shows us:

\(\displaystyle f(2)=2^3-3(2)-3=-1\)

\(\displaystyle f(3)=3^3-3(3)-3=15\)

So, we know $2<a<3$...and since $10^2<127<10^3$, we also know $2<\log(127)<3$. And so, we may begin the continued fraction representation of each by stating:

\(\displaystyle a\approx[2;\cdots]\)

\(\displaystyle \log(127)\approx[2;\cdots]\)

For the purpose of continuing the computation of the continued fraction representation of $\log(127)$, we will use:

\(\displaystyle \log(127)=2+\log(1.27)\)

To get the next partial quotient for $a$, we will make the substitution:

\(\displaystyle x=\frac{1}{x_1}+2\)

and we then find:

\(\displaystyle f_1\left(x_1\right)=x_1^3f\left(\frac{1}{x_1}+2\right)=x_1^3\left(\left(\frac{2x_1+1}{x_1}\right)^3-3\left(\frac{2x_1+1}{x_1}\right)-3\right)=-x_1^3+9x_1^2+6x_1+1\)

We then observe:

\(\displaystyle f_1(9)=55\)

\(\displaystyle f_1(10)=-39\)

And so we have:

\(\displaystyle a\approx[2;9,\cdots]\)

And using the iterative method outlined >>here<<, we confirm that:

\(\displaystyle 1.27^9<10<1.27^{10}\)

Hence:

\(\displaystyle \log(127)\approx[2;9,\cdots]\)

So, we find the next partial quotient for $a$:

\(\displaystyle f_2\left(x_2\right)=x_2^3f_1\left(\frac{9x_2+1}{x_2}\right)=x_2^3\left(-\left(\frac{9x_2+1}{x_2}\right)^3+9\left(\frac{9x_2+1}{x_2}\right)^2+6\left(\frac{9x_2+1}{x_2}\right)+1\right)=55x_2^3-75x_2^2-18x_2-1\)

We find:

\(\displaystyle f_2(1)=-39\)

\(\displaystyle f_2(2)=103\)

Thus:

\(\displaystyle a\approx[2;9,1,\cdots]\)

And we then confirm:

\(\displaystyle \left(\frac{10}{1.27^9}\right)^1<1.27<\left(\frac{10}{1.27^9}\right)^2\)

Hence:

\(\displaystyle \log(127)\approx[2;9,1,\cdots]\)

Continuing this process, we eventually find:

\(\displaystyle a\approx[2;9,1,1,1,2,1,\cdots]\)

\(\displaystyle \log(127)\approx[2;9,1,1,1,2,4,\cdots]\)

And so we may now conclude:

\(\displaystyle a<\log(127)\)

 

[anemone]

Awesome, MarkFL!(Yes)