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Solution 3.4:2c

From Förberedande kurs i matematik 1

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Current revision (11:28, 2 October 2008) (edit) (undo)
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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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Current revision

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,

LHSRHS=ln3ex2=ln3+lnex2=ln3+x2lne=ln3+x2,=ln2x=xln2.

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

x2−xln2+ln3=0

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

x−21ln22−21ln22+ln3=0,x−21ln22=21ln22−ln3.

Now, we have to be cautious and remember that, because 2e3 and thus ln21ln3 we have that 41(ln2)2ln3 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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