Lösung 3.4:2c
Aus Online Mathematik Brückenkurs 1
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Regardless of what value ''x'' has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get ''x'' down from the exponents, | Regardless of what value ''x'' has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get ''x'' down from the exponents, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\text{LHS} | \text{LHS} | ||
&= \ln\bigl( 3e^{x^{2}}\bigr) | &= \ln\bigl( 3e^{x^{2}}\bigr) | ||
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If we collect the terms onto one side, the equation becomes | If we collect the terms onto one side, the equation becomes | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}} |
which is a standard second-order equation for which we complete the square | which is a standard second-order equation for which we complete the square | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt] | \Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt] | ||
\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.} | \Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.} | ||
Version vom 08:46, 22. Okt. 2008
Regardless of what value x has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get x down from the exponents,
| \displaystyle \begin{align}
\text{LHS} &= \ln\bigl( 3e^{x^{2}}\bigr) = \ln 3 + \ln e^{x^2} = \ln 3 + x^2\ln e = \ln 3 + x^2\,\textrm{,}\\[5pt] \text{RHS} &= \ln 2^x = x\ln 2\,\textrm{.} \end{align} |
If we collect the terms onto one side, the equation becomes
| \displaystyle x^{2}-x\cdot \ln 2 + \ln 3 = 0 |
which is a standard second-order equation for which we complete the square
| \displaystyle \begin{align}
\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt] \Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.} \end{align} |
Now, we have to be cautious and remember that, because \displaystyle 2 < e < 3 and thus \displaystyle \ln 2 < 1 < \ln 3 we have that \displaystyle \tfrac{1}{4}(\ln 2)^2 < \ln 3 and the right-hand side is therefore negative. Since the square on the left-hand side cannot be negative, the equation has no solution.
