From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	Both left- and right-hand sides are positive for all values of ''x'' and this means that we can take the logarithm of both sides and get a more manageable equation,
-	<center> [[Bild:3_4_3a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
-	{{NAVCONTENT_START}}	+	\text{LHS} &= \ln 2^{-x^{2}} = -x^{2}\cdot \ln 2\,,\\[5pt]
-	<center> [[Bild:3_4_3a-2(2).gif]] </center>	+	\text{RHS} &= \ln \bigl(2e^{2x}\bigr) = \ln 2 + \ln e^{2x} = \ln 2 + 2x\cdot \ln e = \ln 2 + 2x\cdot 1\,\textrm{.}
-	{{NAVCONTENT_STOP}}	+	\end{align}</math>}}
		+
		+	After a little rearranging, the equation becomes
		+
		+	{{Displayed math||<math>x^{2}+\frac{2}{\ln 2}x+1=0\,\textrm{.}</math>}}
		+
		+	We complete the square of the left-hand side,
		+
		+	{{Displayed math||<math>\Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} - \Bigl(\frac{1}{\ln 2} \Bigr)^{2} + 1 = 0\,,</math>}}
		+
		+	and move the constant terms over to the right-hand side,
		+
		+	{{Displayed math||<math>\Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} = \Bigl(\frac{1}{\ln 2} \Bigr)^{2} - 1\,\textrm{.}</math>}}
		+
		+	It can be difficult to see whether the right-hand side is positive or not, but if we remember that <math>e > 2</math> and that thus <math>\ln 2 < \ln e = 1\,</math>, we must have that <math>(1/\ln 2)^{2} > 1\,</math>, i.e. the right-hand side is positive.
		+
		+	The equation therefore has the solutions
		+
		+	{{Displayed math||<math>x=-\frac{1}{\ln 2}\pm \sqrt{\Bigl(\frac{1}{\ln 2} \Bigr)^{2}-1}\,,</math>}}
		+
		+	which can also be written as
		+
		+	{{Displayed math||<math>x=\frac{-1\pm \sqrt{1-(\ln 2)^{2}}}{\ln 2}\,\textrm{.}</math>}}

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Current revision

Both left- and right-hand sides are positive for all values of x and this means that we can take the logarithm of both sides and get a more manageable equation,

\displaystyle \begin{align} \text{LHS} &= \ln 2^{-x^{2}} = -x^{2}\cdot \ln 2\,,\\[5pt] \text{RHS} &= \ln \bigl(2e^{2x}\bigr) = \ln 2 + \ln e^{2x} = \ln 2 + 2x\cdot \ln e = \ln 2 + 2x\cdot 1\,\textrm{.} \end{align}

After a little rearranging, the equation becomes

\displaystyle x^{2}+\frac{2}{\ln 2}x+1=0\,\textrm{.}

We complete the square of the left-hand side,

\displaystyle \Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} - \Bigl(\frac{1}{\ln 2} \Bigr)^{2} + 1 = 0\,,

and move the constant terms over to the right-hand side,

\displaystyle \Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} = \Bigl(\frac{1}{\ln 2} \Bigr)^{2} - 1\,\textrm{.}

It can be difficult to see whether the right-hand side is positive or not, but if we remember that \displaystyle e > 2 and that thus \displaystyle \ln 2 < \ln e = 1\,, we must have that \displaystyle (1/\ln 2)^{2} > 1\,, i.e. the right-hand side is positive.

The equation therefore has the solutions

\displaystyle x=-\frac{1}{\ln 2}\pm \sqrt{\Bigl(\frac{1}{\ln 2} \Bigr)^{2}-1}\,,

which can also be written as

\displaystyle x=\frac{-1\pm \sqrt{1-(\ln 2)^{2}}}{\ln 2}\,\textrm{.}