I would like to solve the equation: $2\ln(y) = x$
First way: $$2\ln y = x \Leftrightarrow e^{2\ln y}=e^x \Leftrightarrow y^2=e^x\Rightarrow y= \pm \sqrt{e^x}$$ Second way: $$2\ln y = x \Leftrightarrow \ln y= \frac x2 \Leftrightarrow e^{\ln y}=e^{ \frac x2} \Leftrightarrow y=e^{ \frac x2} \Leftrightarrow y= \sqrt{e^x}$$
Why do we get an extra root in the first solution? Which solution is the correct one?
On
Have you already checked your result with a numeric example?
Let's say $x = 2$ and thus your solution for y = -2.717882 and + 2.717882.
but the base of the natural logarithm should always be positive. For a good explanation why, see this post:
Why must the base of a logarithm be a positive real number not equal to 1?
$\ln y$ is defined only for $y>0$ so you have to choose the positive square root in the first method.