Section 1.9 : Exponential And Logarithm Equations
4. Find all the solutions to \(4x + 1 = \left( {12x + 3} \right){{\bf{e}}^{{x^2} - 2}}\). If there are no solutions clearly explain why.
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It may not be apparent at first glance, but with some work we can do a little factoring on this equation. To do that first move everything to one side and then the factoring might become a little more apparent.
\[\begin{align*}4x + 1 - \left( {12x + 3} \right){{\bf{e}}^{{x^2} - 2}} & = 0\\ \left( {4x + 1} \right) - 3\left( {4x + 1} \right){{\bf{e}}^{{x^2} - 2}} & = 0\\ \left( {4x + 1} \right)\left( {1 - 3{{\bf{e}}^{{x^2} - 2}}} \right) & = 0\end{align*}\]Note that in the second step we put parenthesis around the first couple of terms solely to make the factoring in the next step a little more apparent. It does not need to be done in practice.
Be careful to not cancel the \(4x + 1\) from both terms. When solving equations you can only cancel something if you know for a fact that it won’t be zero. If the term can be zero and you cancel it you will miss solutions, and that will be the case here.
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We now have a product of terms that is equal to zero so we know,
\[4x + 1 = 0\hspace{0.5in}{\rm{OR}}\hspace{0.5in}1 - 3{{\bf{e}}^{{x^2} - 2}} = 0\]From the first equation we can quickly arrive at one solution, \(x = - \frac{1}{4}\), and again note that if we had canceled the \(4x + 1\) at the beginning we would have missed this solution. Now all we need to do is solve the equation involving the exponential.
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We can now solve the exponential equation in the same manner as the first couple of problems in this section.
\[\begin{align*}{{\bf{e}}^{{x^2} - 2}} & = \frac{1}{3}\\ \ln \left( {{{\bf{e}}^{{x^2} - 2}}} \right) & = \ln \left( {\frac{1}{3}} \right)\\ {x^2} - 2 & = \ln \left( {\frac{1}{3}} \right)\\ {x^2} & = 2 + \ln \left( {\frac{1}{3}} \right)\\ x & = { \pm \sqrt {2 + \ln \left( {\frac{1}{3}} \right)} = \pm 0.9494}\end{align*}\]Depending upon your preferences either the exact or decimal solution can be used.
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So, we have the following solutions to this equation.
\[ \require{bbox} \bbox[2pt,border:1px solid black]{x = -\frac{1}{4} \hspace{0.25in} {\rm{OR }} \hspace{0.25in} x = \pm \sqrt {2 + \ln \left( {\frac{1}{3}} \right)} = \pm 0.9494} \]