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Section 1-9 : Exponential And Logarithm Equations

7. Find all the solutions to \(\displaystyle 1 - 8\ln \left( {\frac{{2x - 1}}{7}} \right) = 14\). 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 logarithm on one side by itself with a coefficient of one.

\[1 - 8\ln \left( {\frac{{2x - 1}}{7}} \right) = 14\hspace{0.5in} \Rightarrow \hspace{0.5in}\ln \left( {\frac{{2x - 1}}{7}} \right) = - \frac{{13}}{8}\] Show Step 2

Now all we need to do is exponentiate both sides using \(\bf{e}\) (because we’re working with the natural logarithm) and then solve for \(x\).

\[\begin{align*}{{\bf{e}}^{\ln \left( {\frac{{2x - 1}}{7}} \right)}} & = {{\bf{e}}^{ - \,\frac{{13}}{8}}}\\ \frac{{2x - 1}}{7} & = {{\bf{e}}^{ - \,\frac{{13}}{8}}}\\ x & = \frac{1}{2}\left( {1 + 7{{\bf{e}}^{ - \frac{{13}}{8}}}} \right) & = 1.1892\end{align*}\] Show Step 3

We’re dealing with logarithms so we need to make sure that we won’t have any problems with any of our potential solutions. In other words, we need to make sure that if we plug in the potential solution into the original equation we won’t end up taking the logarithm of a negative number or zero.

Plugging in we can see that we won’t be taking the logarithm of a negative number and so the solution is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{1}{2}\left( {1 + 7{{\bf{e}}^{ - \,\frac{{13}}{8}}}} \right) = 1.1892}}\]

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