# Solving the equation $x^{ (\frac{x}{123}+11)} =123$

I came across this equation

$$x^{ \left(\frac{x}{123} + 11 \right) } = 123$$

All I could think of is to put $\ln$ into the equation:

\begin{align} \ln\left(x^{ \left( \frac{x}{123} + 11 \right) } \right) &= \ln\left(123 \right) \\ \ln(x)\cdot \left(\dfrac{x}{123}+11\right) &= \ln\left(123 \right) \end{align}

and I'm lost.

What should I do now?

-
This doesn't look solvable, but it appears that $x \in (1,2)$ –  Rustyn Mar 4 '13 at 5:52
@RustynYazdanpour Sorry I wrote the wrong title... It is supposed to be $x^{x/123+11}=123$ –  User 2.71 Mar 4 '13 at 5:56
WolframAlpha says $x \approx 1.548$. Are you supposed to approximate it? –  Michael Biro Mar 4 '13 at 5:58
@MichaelBiro Yeah. And mainly I want to know what algorithm should be applied to question of this type –  User 2.71 Mar 4 '13 at 6:00
You were lost before that. Taking $\ln$ of both sides you should get $\left(\frac{x}{123}+11\right) \ln x = \ln 123$. –  Robert Israel Mar 4 '13 at 6:09

As @MichaelBiro states above in the comments, wolframalpha, gives an approximate solution over the reals: $$x\approx 1.54801$$ Using this value of $x$ gives a relative error of: $$\approx -.001 \text{%}$$ You can try Newton's method, since you know your root $x^* \in (1,2)$
The newton iteration is as follows: $$x_0 = 1.5 \\$$ $$x_{n+1} = x_n-\frac{(123 x_n^{\left(\tfrac{-x_n}{123}-10\right)})(x_n^{\left(\tfrac{x_n}{123}+11\right)}-123)}{(x (\ln(x)+1)+1353)}$$