Solution 3.4:3a

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