Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

2 Answers 2

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

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.