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I need the value of $x$, but I cannot figure out how to deal with ${\sqrt x-1}$ in the following equation.

$x^{\sqrt x-1} =3/2$

Could anyone please help me to figure this out.

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    $\begingroup$ You need numerical methods (or perhaps the Lambert-W-function can be used) $\endgroup$
    – Peter
    Dec 28, 2020 at 10:34
  • $\begingroup$ @Peter I'm not familiar with numerical method. My mathematics knowledge is of elementary level. Could you please suggest something a elementary student should know? $\endgroup$
    – Russell
    Dec 28, 2020 at 10:55

4 Answers 4

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x=9/4 is a solution

$\sqrt(9/4)=3/2$

$3/2-1=1/2$

$9/4^{\sqrt(9/4)-1}=\sqrt(9/4)=3/2$

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  • $\begingroup$ Have you calculated it backwards? $\endgroup$
    – Russell
    Dec 28, 2020 at 11:34
  • $\begingroup$ It was kind of obvious. $\endgroup$
    – yugikaiba
    Dec 28, 2020 at 11:35
  • $\begingroup$ Because you're a genius. But I am having a really hard time to understand how it works. $\endgroup$
    – Russell
    Dec 28, 2020 at 11:41
  • $\begingroup$ I was lucky and from practice but if something seems hard is because there might be some "lucky" idea behind. Next one will be easier to find. $\endgroup$
    – yugikaiba
    Dec 28, 2020 at 11:44
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Let's substitute, $x=y^2$,

$$ x^{\sqrt{x}-1}=(y^2)^{y-1} = y^{2y-2} = \frac{3}{2} $$

$$ \implies y^{y-1} = \sqrt{\frac{3}{2}} $$

This is still an implicit equation but we got rid of the $\sqrt{x}$.

Here is the Numerical Solution by Wolfram. This problem cannot be solved analytically using elementary operations. You need to follow some algorithm like [Bisection Method][2]. They work by taking some initial interval at the start say $[1, 100]$. Check the function value at both points. Then bisect the current interval i.e. check the value at $y=50$. We see that the root lies in $[1, 50]$. This way we go on bisecting the interval to get to the solution.

The solution using any of the algorithms is $x=0.3616$ or $x=2.25$.

[2]: https://en.wikipedia.org/wiki/Root-finding_algorithms#:~:text=The%20simplest%20root%2Dfinding%20algorithm,point%20that%20bisects%20the%20interval).

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    $\begingroup$ Sorry, I should have been a little more specific in my question. I need the value of $x$. $\endgroup$
    – Russell
    Dec 28, 2020 at 10:50
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Consider that you are looking for the zero's of function $$f(x)=x^{\sqrt x-1} -\frac 32$$ Its derivatives are $$f'(x)=\frac{1}{2} x^{\sqrt{x}-2} \left(2 \sqrt{x}+\sqrt{x} \log (x)-2\right)$$ $$f''(x)=\frac{1}{4} x^{\sqrt{x}-3} \left(4 \left(x-2 \sqrt{x}+2\right)+\log (x) \left(4 x-5 \sqrt{x}+x \log (x)\right)\right)$$

The first derivative cancels at $x=1$ and $f(1)=-\frac{1}{2}$, $f''(1)=1$. So, this point is a minimum and there are two solutions $$0 < x_1 < 1 \quad 1 < x_2$$

Build a Taylor series around $x=1$. It will be $$-\frac{1}{2}+\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$ This gives two solutions : $0$ and $2$ and $0$ must be discarded.

$$f(2)=2^{\sqrt{2}-1}-\frac{3}{2}\sim -0.167428 \implies x_2 >2$$ and you were already told that $x=\frac 94$ is the largest solution. Then, I shall not focus on this one.

Adding one more term to the expansion, we have $$-\frac{1}{2}+\frac{1}{2} (x-1)^2-\frac{3}{8} (x-1)^3+O\left((x-1)^4\right)$$ which is a cubic in $(x-1)$. The discriminant being negative $(\Delta=-\frac{179}{256})$, there is only one real root. Using the hyperbolic method, we have $$\frac{1}{9} \left(13-8 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{211}{32}\right)\right)\right)\sim 0.207953$$

So, now we have our starting guess for Newton method. For the smallest root, the iterates are $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.207953 \\ 1 & 0.291319 \\ 2 & 0.346314 \\ 3 & 0.360882 \\ 4 & 0.361632 \\ 5 & 0.361634 \end{array} \right)$$

This number is not recognized by inverse symbolic calculators, but (given for the fun) it is close to $$\frac{11+\sqrt{23}}{50}\, \sqrt[3]{\frac{3}{2}} \sim 0.361634266$$ while the "exact" solution is $0.361634272$.

Even if this does not mean any thing, for this funny number $f(x)=1.56\times 10^{-8}$.

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I performed an iterative solution with an Android phone simulator of an HP-$42$S. I started by taking the log of both sides. $$x^{\sqrt{x}-1}=\frac{3}{2}$$ $$(\sqrt{x}-1)\cdot\log(x)=\log\left(\frac{3}{2}\right)$$ $$\sqrt{x}-1=\frac{\log\left(\frac{3}{2}\right)}{\log(x)}$$ $$\sqrt{x}=\frac{\log\left(\frac{3}{2}\right)}{\log(x)}+1$$ $$x=\left(\frac{\log\left(\frac{3}{2}\right)}{\log(x)}+1\right)^2$$ Start with an initial value of $x=5$ and watch a solution evolve.

After coding the below, plug in 5, and XEG "SME". The HP-$42$S code follows: LBL "SME" STO 01 LBL 01 3 2 / LOG RCL 01 LOG / 1 + X^2 STO 01 VIEW ST X PSE GTO 01

This is at least one solution there may be others.

The simulator is so fast, if you take the PSE out, you will just see a solution. For example, incrementing X from 0 to 1,000,000 takes seconds.

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