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Section 1-9 : Exponential And Logarithm Equations

3. Find all the solutions to \(2t - t{{\bf{e}}^{6\,t - 1}} = 0\). 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

First notice that we can factor a \(t\) out of both terms to get,

\[t\left( {2 - {{\bf{e}}^{6\,t - 1}}} \right) = 0\]

Be careful to not cancel the \(t\) 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 the case here.

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We now have a product of terms that is equal to zero so we know,

\[t = 0\hspace{0.5in}{\rm{OR}}\hspace{0.5in}2 - {{\bf{e}}^{6\,t - 1}} = 0\]

So, we have one solution already, \(t = 0\), and again note that if we had canceled the \(t\) 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.

\[\begin{align*}{{\bf{e}}^{6\,t - 1}} & = 2\\ \ln \left( {{{\bf{e}}^{6\,t - 1}}} \right) & = \ln \left( 2 \right)\\ 6t - 1 & = \ln \left( 2 \right)\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{6}\left( {1 + \ln \left( 2 \right)} \right) = 0.2822}}\end{align*}\]

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