# an exponent problem

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

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

• You need numerical methods (or perhaps the Lambert-W-function can be used) Dec 28, 2020 at 10:34
• @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? Dec 28, 2020 at 10:55

x=9/4 is a solution

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

$$3/2-1=1/2$$

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

• Have you calculated it backwards? Dec 28, 2020 at 11:34
• It was kind of obvious. Dec 28, 2020 at 11:35
• Because you're a genius. But I am having a really hard time to understand how it works. Dec 28, 2020 at 11:41
• 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. Dec 28, 2020 at 11:44

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 $$x=2.25$$.

• Sorry, I should have been a little more specific in my question. I need the value of $x$. Dec 28, 2020 at 10:50

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 $$f(x)=1.56\times 10^{-8}$$.

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.