# Terrible equation $(x+x^{\ln x})^{10}=2^{10}$

$$(x+x^{\ln x})^{10}=2^{10}$$ 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?

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

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

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:

\begin{array}{c|c}n&x_n\\\hline0&0.2\\1&0.253997\\2&0.322907\\3&0.402142\\4&0.476180\\5&0.523933\\6&0.539434\\7&0.540775\\8&0.540784\\9&0.540784\end{array}

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.

• Personally, I like to show the process of the iteration, but good answer anyways :D – Simply Beautiful Art Jan 24 '17 at 22:24
• @SimplyBeautifulArt I am actually adding this process to my answer now. – projectilemotion Jan 24 '17 at 22:25
• Oh. :D (and just for the record, I like tables like this: math.stackexchange.com/questions/2112441/…) – Simply Beautiful Art Jan 24 '17 at 22:26
• Nice job. +2 :D – Simply Beautiful Art Jan 24 '17 at 22:43
• @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? – Denis Nichita Jan 25 '17 at 16:52