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

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The discussion revolves around proving that for the real root \( a \) of the equation \( x^3 - 3x - 3 = 0 \), the inequality \( 10^a < 127 \) holds true. The original poster acknowledges a mistake in their initial solution attempt and expresses frustration over not being able to solve the problem as intended. They plan to continue working on the problem and consider alternative methods suggested by other participants. The conversation highlights the collaborative nature of problem-solving in mathematics. Ultimately, the focus remains on finding a valid proof for the inequality involving the real root \( a \).
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Given $a$ is the real root for $x^3-3x-3=0$, prove that $10^a\lt 127$.
 
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anemone said:
Given $a$ is the real root for $x^3-3x-3=0$, prove that $10^a\lt 127$.
my solution:
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 $log 127\approx 2.10381$
$\therefore 10^a<127$
 
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anemone said:
Given $a$ is the real root for $x^3-3x-3=0$, prove that $10^a\lt 127$.

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

...but I don't have proof. :(
 
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):o

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.
 
My solution:

It will suffice to show that:

$$a<\log(127)$$

If we define:

$$f(x)=x^3-3x-3$$

A quick sign check shows us:

$$f(2)=2^3-3(2)-3=-1$$

$$f(3)=3^3-3(3)-3=15$$

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:

$$a\approx[2;\cdots]$$

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

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

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

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

$$x=\frac{1}{x_1}+2$$

and we then find:

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

We then observe:

$$f_1(9)=55$$

$$f_1(10)=-39$$

And so we have:

$$a\approx[2;9,\cdots]$$

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

$$1.27^9<10<1.27^{10}$$

Hence:

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

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

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

We find:

$$f_2(1)=-39$$

$$f_2(2)=103$$

Thus:

$$a\approx[2;9,1,\cdots]$$

And we then confirm:

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

Hence:

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

Continuing this process, we eventually find:

$$a\approx[2;9,1,1,1,2,1,\cdots]$$

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

And so we may now conclude:

$$a<\log(127)$$
 
Awesome, MarkFL!(Yes):cool:
 
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