$$(x+x^{\ln x})^{10}=2^{10}$$ I tried to take the logarithm of both sides. To apply 10th root, but didn't get too far. Please help!

Edit: I have the following result which tells me that it has only one solution. $$x^{\ln x}=2-x$$ we take logarithm on both sides$$\ln^2{x}=\ln{(2-x)}$$ $\ln^2{x}$ is strictly increasing function (for $x>0$) and this function $\ln(2-x)$ is strictly decreasing (because is a composition of a strictly increasing one with a decreasing one). So we have only one solution $x=1$. What is wrong here?

  • 1
    $\begingroup$ $x=1$ is a solution. $\endgroup$ – Adren Jan 24 '17 at 22:03
  • $\begingroup$ It has another solution besides $1$ as well.. $\endgroup$ – MathematicianByMistake Jan 24 '17 at 22:04
  • $\begingroup$ You got as far as $|x+x^{\ln x}| = 2$, right? And $x$ must be positive, otherwise $\ln x$ is undefined; so $x+x^{\ln x} = 2$. $\endgroup$ – TonyK Jan 24 '17 at 22:06
  • $\begingroup$ There is another solution, slightly greater than $1/2$. $\endgroup$ – Adren Jan 24 '17 at 22:08
  • $\begingroup$ @TonyK Yess. That far I went $\endgroup$ – Denis Nichita Jan 24 '17 at 22:12

After taking the 10th root on both sides, we obtain: $$|x+x^{\ln{x}}|=2$$ Note that since $x \in \mathbb{R}^+$, we can deduce that: $$x+x^{\ln{x}}=2$$ Now, this equation has a solution which is obvious, at $x=1$. However, there is another solution to this equation.

I do not think there is a closed form solution in terms of elementary functions for $x$. Therefore, we must use a numerical method. I will use the Newton-Raphson Method.

The process is as follows:


We choose an initial starting point $x_0=0.2$, a reasonable estimate of the solution.

We use the functions:

$f(x)=x+x^{\ln x}-2$

And find its derivative:

$f'(x)=2\ln{x} \cdot x^{\ln{x}-1}+1$

To obtain the iteration:

$$x_{n+1}=x_n-\frac{x_n+{x_n}^{\ln{x_n}}-2}{2\ln{x_n} \cdot {x_n}^{\ln{x_n}-1}+1}$$

Doing this gives the solution:


As the iterations $n \to \infty$, $x_n \to x$, the solution to the equation. Therefore:

$$x \approx 0.54078414712$$

You can implement this method to a spreadsheet, or by a more sophisticated software such as MATLAB.

  • $\begingroup$ Personally, I like to show the process of the iteration, but good answer anyways :D $\endgroup$ – Simply Beautiful Art Jan 24 '17 at 22:24
  • $\begingroup$ @SimplyBeautifulArt I am actually adding this process to my answer now. $\endgroup$ – projectilemotion Jan 24 '17 at 22:25
  • $\begingroup$ Oh. :D (and just for the record, I like tables like this: math.stackexchange.com/questions/2112441/…) $\endgroup$ – Simply Beautiful Art Jan 24 '17 at 22:26
  • 1
    $\begingroup$ Nice job. +2 :D $\endgroup$ – Simply Beautiful Art Jan 24 '17 at 22:43
  • $\begingroup$ @projectilemotion with the method you used, how do you know which solution your sequence aproximates? I presume this method could be used to aproximate also $x=1$. What starting point do you get? $\endgroup$ – Denis Nichita Jan 25 '17 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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