Lösung 3.4:2c

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Regardless of what value
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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,
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<math>x</math>
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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
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<math>x</math>
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down from the exponents:
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LHS
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<math>=\ln \left( 3e^{x^{2}} \right)=\ln 3+\ln e^{x^{2}}=\ln 3+x^{2}\ln e=\ln 3+x^{2}</math>
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RHS
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<math>=\ln 2^{x}=x\ln 2</math>
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{{Displayed math||<math>\begin{align}
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\text{LHS}
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&= \ln\bigl( 3e^{x^{2}}\bigr)
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= \ln 3 + \ln e^{x^2}
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= \ln 3 + x^2\ln e
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= \ln 3 + x^2\,\textrm{,}\\[5pt]
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\text{RHS}
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&= \ln 2^x = x\ln 2\,\textrm{.}
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\end{align}</math>}}
If we collect the terms onto one side, the equation becomes
If we collect the terms onto one side, the equation becomes
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{{Displayed math||<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}
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<math>x^{2}-x\centerdot \ln 2+\ln 3=0</math>
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which is a standard second-order equation for which we complete the square
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which is a standard second-order equation for which we complete the square:
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \left( x-\frac{1}{2}\ln 2 \right)^{2}-\left( \frac{1}{2}\ln 2 \right)^{2}+\ln 3=0 \\
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\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt]
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& \left( x-\frac{1}{2}\ln 2 \right)^{2}=\left( \frac{1}{2}\ln 2 \right)^{2}-\ln 3 \\
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\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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Now, we have to be cautious and remember that, because
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Now, we have to be cautious and remember that, because <math>2 < e < 3</math> and thus <math>\ln 2 < 1 < \ln 3</math> we have that <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math> and the right-hand side is therefore negative. Since the square on the left-hand side cannot be negative, the equation has no solution.
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<math>\text{2}<e<\text{3}</math>
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and thus
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<math>\text{ln 2}<\text{1}<\text{ln 3}</math>
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which means that
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<math>\frac{1}{4}\left( \ln 2 \right)^{2}<\ln 3</math>
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and the right-hand side is therefore negative. Since the square on the right-hand side cannot be negative, the equation has no solution.
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Version vom 11:28, 2. 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,

Vorlage:Displayed math

If we collect the terms onto one side, the equation becomes

Vorlage:Displayed math

which is a standard second-order equation for which we complete the square

Vorlage:Displayed math

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.

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