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?
 A: 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.
