Lösung 3.4:2b
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| Zeile 1: | Zeile 1: | ||
If we write the equation as | If we write the equation as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\bigl(e^{x}\bigr)^{2} + e^{x} = 4</math>}} |
we see that <math>x</math> appears only in the combination <math>e^{x}</math> and it is therefore appropriate to treat <math>e^{x}</math> as a new unknown in the equation and then, when we have obtained the value of <math>e^{x}</math>, we can calculate the corresponding value of <math>x</math> by simply taking the logarithm. | we see that <math>x</math> appears only in the combination <math>e^{x}</math> and it is therefore appropriate to treat <math>e^{x}</math> as a new unknown in the equation and then, when we have obtained the value of <math>e^{x}</math>, we can calculate the corresponding value of <math>x</math> by simply taking the logarithm. | ||
| Zeile 7: | Zeile 7: | ||
For clarity, we set <math>t=e^{x}</math>, so that the equation can be written as | For clarity, we set <math>t=e^{x}</math>, so that the equation can be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>t^{2}+t=4</math>}} |
and we solve this second-degree equation by completing the square, | and we solve this second-degree equation by completing the square, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>t^{2}+t = \Bigl( t+\frac{1}{2} \Bigr)^{2}-\Bigl( \frac{1}{2} \Bigr)^{2} = \Bigl( t+\frac{1}{2} \Bigr)^{2} - \frac{1}{4}\,,</math>}} |
which gives | which gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\Bigl(t+\frac{1}{2}\Bigr)^{2} - \frac{1}{4} = 4\quad \Leftrightarrow \quad t = -\frac{1}{2}\pm \frac{\sqrt{17}}{2}\,\textrm{.}</math>}} |
These two roots give us two possible values for <math>e^{x}</math>, | These two roots give us two possible values for <math>e^{x}</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}\qquad\text{or}\qquad e^{x} = -\frac{1}{2}+\frac{\sqrt{17}}{2}\,\textrm{.}</math>}} |
In the first case, the right-hand side is negative and because "''e'' raised to anything" can never be negative, there is no ''x'' that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because | In the first case, the right-hand side is negative and because "''e'' raised to anything" can never be negative, there is no ''x'' that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because | ||
<math>\sqrt{17}>1</math>) and we can take the logarithm of both sides to obtain | <math>\sqrt{17}>1</math>) and we can take the logarithm of both sides to obtain | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x=\ln \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\,\textrm{.}</math>}} |
Note: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting <math>t=\sqrt{17}/2-1/2</math> into the equation <math>t^2+t=4</math>, | Note: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting <math>t=\sqrt{17}/2-1/2</math> into the equation <math>t^2+t=4</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\text{LHS} | \text{LHS} | ||
&= \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)^2 + \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\\[5pt] | &= \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)^2 + \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\\[5pt] | ||
Version vom 08:46, 22. Okt. 2008
If we write the equation as
| \displaystyle \bigl(e^{x}\bigr)^{2} + e^{x} = 4 |
we see that \displaystyle x appears only in the combination \displaystyle e^{x} and it is therefore appropriate to treat \displaystyle e^{x} as a new unknown in the equation and then, when we have obtained the value of \displaystyle e^{x}, we can calculate the corresponding value of \displaystyle x by simply taking the logarithm.
For clarity, we set \displaystyle t=e^{x}, so that the equation can be written as
| \displaystyle t^{2}+t=4 |
and we solve this second-degree equation by completing the square,
| \displaystyle t^{2}+t = \Bigl( t+\frac{1}{2} \Bigr)^{2}-\Bigl( \frac{1}{2} \Bigr)^{2} = \Bigl( t+\frac{1}{2} \Bigr)^{2} - \frac{1}{4}\,, |
which gives
| \displaystyle \Bigl(t+\frac{1}{2}\Bigr)^{2} - \frac{1}{4} = 4\quad \Leftrightarrow \quad t = -\frac{1}{2}\pm \frac{\sqrt{17}}{2}\,\textrm{.} |
These two roots give us two possible values for \displaystyle e^{x},
| \displaystyle e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}\qquad\text{or}\qquad e^{x} = -\frac{1}{2}+\frac{\sqrt{17}}{2}\,\textrm{.} |
In the first case, the right-hand side is negative and because "e raised to anything" can never be negative, there is no x that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because \displaystyle \sqrt{17}>1) and we can take the logarithm of both sides to obtain
| \displaystyle x=\ln \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\,\textrm{.} |
Note: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting \displaystyle t=\sqrt{17}/2-1/2 into the equation \displaystyle t^2+t=4,
| \displaystyle \begin{align}
\text{LHS} &= \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)^2 + \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\\[5pt] &= \frac{17}{4}-2\cdot \frac{1}{2}\cdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2}\\[5pt] &= \frac{17}{4}+\frac{1}{4}-\frac{1}{2}\\[5pt] &= \frac{17+1-2}{4}\\[5pt] &=\frac{16}{4}\\[5pt] &= 4\\[5pt] &= \text{RHS.} \end{align} |
