+29 I believe in this case x can't be a whole number which means you will have to go through some trial and errors.
when x is 1.5, x^x=1.8371173070873836.... which is obviously smaller than 2
when x=1.6 x^x=2.1212505710975916.... which is bigger than 2
now we know that x is between 1.5 and 1.6
you can always write "x is bigger than 1.5 but smaller than 1.6" or you can do some trial and errors to get more accurate answer such as x=1.559610.....
but i believe you won't be able to get an accurate answer as it is an infinite decimal.
hope it helped :)
Taking logs, the equation can be written as 
,
and this requires a numerical method for its solution, Newton-Raphson being favourite.
A simpler possibility, but having a slower rate of convergence than Newton-Raphson, is to write the equation as
from which
Now choose a starter value for x, substitute that into the rhs to calculate a new value for x and repeat that until you have the required degree of accuracy.
Starting with x = 2, the sequence is 1.4142, 1.6325, 1.5290, 1.5736, 1.5535, 1.5624,...
but you could use a different starting value.
+26393 x^x=2. How to solve this equation?
$$\small{\text{$
\begin{array}{rcl}
x^x &=&2 \qquad | \qquad \ln{()}\\
x\ln(x) &=& \ln{(2)} \qquad | \qquad x = e^{ln{(x)}}\\
e^{ln{(x)}}\cdot \ln{(x)} & =& \ln{(2)} \qquad | \qquad z = \ln{(x)} \\
e^z\cdot z &=& \ln{(2)}\\
\end{array}
$}}\\\\
\small{\text{
If we have $\boxed{~a=z\cdot e^z~}$,
the solution is $\boxed{~z=W(a)~}$. and $\boxed{~x=e^z~} $}}\\
\small{\text{
In mathematics, the Lambert W function, is also called the omega function or product logarithm
}}$$
$$\small{\text{$
\begin{array}{lrcl}
\text{We have }& z &=& \ln{(x)}\\
\text{and } & e^z\cdot z &=& \ln{(2)}\\
\text{so } & z &=& W(\ln{(2)})\\
\text{and } & \ln{(x)}&=& W(\ln{(2)})\\\\
& e^{\ln{(x)}} &=& e^{W(\ln{(2)})}\\
\text{finally }& x &=& e^{W(\ln{(2)})}\\
\text{also } & \mathbf{ x }& \mathbf{=} & \mathbf{\dfrac{\ln{(2)}}{W(\ln{(2)})} }
\end{array}
$}}\\\\$$
$$\small{\text{ Numerical Evaluation:}}\\ \small{\text{ The $W$ function may be approximated using Newton's method, with successive approximations to}}\\ \small{\text{ $z=W(a)$ (so $a=ze^z$) being $z_{j+1}=z_j-\dfrac{z_j e^{z_j}-a}{e^{z_j}+z_j e^{z_j}}$. }}\\
\small{\text{We start the Iteration with $z_0 = 1$}}\\$$
$$\small{\text{$
\begin{array}{rclcrcl}
z_1 &=& 1 - \dfrac{ 1\cdot e^1 - \ln{(2)} }{e^1+ 1\cdot e^1 } = 0.62749729872 &\quad & x_1 &=& \dfrac{ \ln{(2)} } {z_1} = 1.10462177603\\\\
z_2 &=& z_1-\dfrac{z_1 e^{z_1}- \ln{(2)} }{e^{z_1}+z_1 e^{z_1}} = 0.24045491796 &\quad & x_2 &=& \dfrac{ \ln{(2)} } {z_2} = 2.88264921519\\\\
z_3 &=& z_2-\dfrac{z_2 e^{z_2}- \ln{(2)} }{e^{z_2}+z_2 e^{z_2}} = 0.48596644327 &\quad & x_3 &=& \dfrac{ \ln{(2)} } {z_3} = 1.42632724987\\\\
z_4 &=& z_3-\dfrac{z_3 e^{z_3}- \ln{(2)} }{e^{z_3}+z_3 e^{z_3}} = 0.44585119296 &\quad & x_4 &=& \dfrac{ \ln{(2)} } {z_4} = 1.55466036989\\\\
z_5 &=& z_4-\dfrac{z_4 e^{z_4}- \ln{(2)} }{e^{z_4}+z_4 e^{z_4}} = 0.44443778365 &\quad & x_5 &=& \dfrac{ \ln{(2)} } {z_5} = 1.55960452971\\\\
z_6 &=& z_5-\dfrac{z_5 e^{z_5}- \ln{(2)} }{e^{z_5}+z_5 e^{z_5}} = 0.44443609102 &\quad & x_6 &=& \dfrac{ \ln{(2)} } {z_6} = 1.55961046945\\\\
z_7 &=& z_6-\dfrac{z_6 e^{z_6}- \ln{(2)} }{e^{z_6}+z_6 e^{z_6}} = 0.44443609102 &\quad & x_7 &=& \dfrac{ \ln{(2)} } {z_7} = 1.55961046946\\\\
\end{array}
$}}$$
x = 1.55961046946
