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Looking for some help with trying to solve for $x$ in the following equation,

$$0=5-re^{-x^2}$$

The solution is,

$$x^2=\ln\frac{r}{5}$$

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    Okay, and what's your question? – 2017-02-14
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    I am just looking for some help with the steps inbetween – 2017-02-14
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    This seems pretty straight forward,try to adjust the equation and take $\ln$.Also this question has nothing to do with system of equations. – 2017-02-14
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    We must have r>=5. – 2017-02-14

2 Answers 2

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\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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    does the fraction of the ln switch because of the negative sign causing the ln5 to be negative? – 2017-02-14
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    It's a property of logarithms. See my edit. – 2017-02-14
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There you go

0=5−re-x2

e-x2=5/r

ln(e-x2)=ln(5/r)

-x2=ln(5/r)

x2=-ln(5/r)

x2=ln[(5/r)-1]

x2=ln(r/5)