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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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Hint : Be careful to not cancel terms that shouldn’t be canceled. Remember that you can’t cancel something unless you know for a fact that it won’t ever be zero. Also, note that if you can cancel something then it can be factored out of the equation.
Start Solution

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.

Show Step 3

We can now solve the exponential equation in the same manner as the first couple of problems in this section.

\[\begin{align*}{{\bf{e}}^{{x^2} - 2}} & = \frac{1}{3}\\ \ln \left( {{{\bf{e}}^{{x^2} - 2}}} \right) & = \ln \left( {\frac{1}{3}} \right)\\ {x^2} - 2 & = \ln \left( {\frac{1}{3}} \right)\\ {x^2} & = 2 + \ln \left( {\frac{1}{3}} \right)\\ x & = \require{bbox} \bbox[2pt,border:1px solid black]{{ \pm \sqrt {2 + \ln \left( {\frac{1}{3}} \right)} = \pm 0.9494}}\end{align*}\]

Depending upon your preferences either the exact or decimal solution can be used.