<s>[INST] Solving for varible with use of ln Looking for some help with trying to solve for $x$ in the following equation,
$$0=5-re^{-x^2}$$
The solution is,
$$x^2=\ln\frac{r}{5}$$
 [/INST] \begin{align*}
\implies 5 = re^{-x^2} \iff \frac{5}{r} = e^{-x^{2}} \iff \ln\frac{5}{r} = -x^{2} \iff -\ln\frac{5}{r} = x^2 \iff \ln \frac{r}{5} = x^{2}
\end{align*} 
Note: $\ln a^{n} = n\ln a$. In this case, $n=-1$, i.e. $\ln a^{-1} = -\ln a$.
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