Lösung 3.4:3a
Aus Online Mathematik Brückenkurs 1
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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, | 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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\text{LHS} &= \ln 2^{-x^{2}} = -x^{2}\cdot \ln 2\,,\\[5pt] | \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{.} | \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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After a little rearranging, the equation becomes | After a little rearranging, the equation becomes | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x^{2}+\frac{2}{\ln 2}x+1=0\,\textrm{.}</math>}} |
We complete the square of the left-hand side, | We complete the square of the left-hand side, | ||
| - | {{ | + | {{Abgesetzte Formel||<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, | and move the constant terms over to the right-hand side, | ||
| - | {{ | + | {{Abgesetzte Formel||<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. | 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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The equation therefore has the solutions | The equation therefore has the solutions | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x=-\frac{1}{\ln 2}\pm \sqrt{\Bigl(\frac{1}{\ln 2} \Bigr)^{2}-1}\,,</math>}} |
which can also be written as | which can also be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x=\frac{-1\pm \sqrt{1-(\ln 2)^{2}}}{\ln 2}\,\textrm{.}</math>}} |
Version vom 08:46, 22. Okt. 2008
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{.} |
