How to solve $\ln(x^2)=x$? I am struggling to solve the equation $\ln(x^2)=x.$ I can't figure out how to isolate the variable. Exponentiating both sides leaves me with $x=e^{\frac{x}{2}}.$ I would really appreciate any guidance, or first steps.
Also, isn't one of the properties of logs that $\ln(x^2)=2\ln(x)?$ How can these two things be equal if the term on the left hand side can input negative numbers, but the term on the right cannot?

Consider that you look for the zero's of function
$$f(x)=\log(x^2)-x$$ for which
$$f'(x)=\frac 2 x-1 \qquad \text{and} \qquad f''(x)=-\frac 2 {x^2} < 0 \qquad \forall x$$
The first derivative cancels at $x=2$ and $f(2)=\log(4)-2 <0$. By the second derivative test $x=2$ corresponds to a maximum; then, no solution for $x >0$.
By inspection, $f(-1)=1$ making the solution to be between $-1$ and $0$.
Being lazy, build the Taylor series around $x=-1$ to get
$$f(x) \sim 1-3 (x+1)-(x+1)^2+O\left((x+1)^3\right)$$ Solving the quadratic and keeping the  solution in the range, then  $x=\frac{1}{2} \left(\sqrt{13}-5\right)\approx -0.697224$ which is not too bad.
For better approximations, instead of Taylor series, build $[2,n]$ Padé approximants and set the numerator equal to zero. For example, for $n=2$,
$$f(x)\sim \frac{1-\frac{23 }{6}(x+1)+\frac{14}{9} (x+1)^2 } {1-\frac{5 }{6}(x+1)+\frac{1}{18} (x+1)^2 }\implies x=-\frac{3 \sqrt{305}-13}{56}\approx -0.703442$$ while the exact solution given by Lambert function would be $\approx -0.703467$.
Using the $[2,4]$ Padé approximant would lead to the exact result (for six significant figures).
Edit
Using @Quanto's answer, we can make amazing approximations building the $[1,n]$ Padé approximant of function $(e^x-x^2)$ around $x=-1$. The problem reduces to a linear equation in $(x+1)$. Below are given the expressions and their decimal values
$$\left(
\begin{array}{ccc}
 n & \text{formula} & \text{value} \\
 0 &  -\frac{2+e}{1+2 e} &    -0.73304361 \\
 1 &  -\frac{3+13 e+2 e^2}{1+11 e+6 e^2} &    -0.70599412 \\
 2 &  -\frac{8+140 e+164 e^2+12 e^3}{2+80 e+194 e^2+48    e^3} & -0.70361456 \\
 3 &  -\frac{30+1674 e+6426 e^2+3390 e^3+144 e^4}{6+738 e+5058    e^2+5142 e^3+720 e^4} & -0.70346790
\end{array}
\right)$$
