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### Section 1-9 : Exponential And Logarithm Equations

1. Find all the solutions to $$12 - 4{{\bf{e}}^{7 + 3\,x}} = 7$$. If there are no solutions clearly explain why.

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There isn’t all that much to do here for this equation. First, we need to isolate the exponential on one side by itself with a coefficient of one.

$- 4{{\bf{e}}^{7 + 3\,x}} = - 5\hspace{0.5in} \Rightarrow \hspace{0.5in}{{\bf{e}}^{7 + 3\,x}} = \frac{5}{4}$ Show Step 2

Now all we need to do is take the natural logarithm of both sides and then solve for $$x$$.

\begin{align*}\ln \left( {{{\bf{e}}^{7 + 3\,x}}} \right) & = \ln \left( {\frac{5}{4}} \right)\\ 7 + 3x & = \ln \left( {\frac{5}{4}} \right)\\ x & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}\left( {\ln \left( {\frac{5}{4}} \right) - 7} \right) = - 2.25895}}\end{align*}

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