I assume this equation is . You can't solve it exactly, you have to use a numerical method.
I will admit that I would have interpreted the question as "if x= 1, what is ", but solving the equation is much more interesting.
Of course, so is the same as and that is the same as . Let y -ln(1.5)x so that and the equation becomes or .
Now, we have y= W(-ln(1.5)) where W is "Lambert's W function", the inverse function to [itex]f(x)= xe^x[/itex].
Finally, then .
Of course, as earboth says, x is not a real number because the domain of the real valued W function, while it includes some negative numbers, does not include any that negative.
Of course, you could use any number of methods of approximation. I tried the risky approach of blind iteration, using the following steps (having drawn the graphs first for a rough idea of the roots, obviously):
We get the solution at approximately the 9th iteration. However I was trying to get the other solution! You could try a slightly better approach, such as interval bisection, Newton-Raphson or linear interpolation or something, but then you probably know all this already.