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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?

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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
up vote 0 down vote accepted

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: $$ x_0 = 1.5 \\ $$ $$ x_{n+1} = x_n-\frac{(123 x_n^{\left(\tfrac{-x_n}{123}-10\right)})(x_n^{\left(\tfrac{x_n}{123}+11\right)}-123)}{(x (\ln(x)+1)+1353)} $$

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You won't find a closed-form solution. Try Newton's method.

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