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Section 6.4 : Solving Logarithm Equations

8. Solve the following equation.

\[\ln \left( {x - 1} \right) = 1 + \ln \left( {3x + 2} \right)\]

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Hint : If we can reduce all the logarithms to a single logarithm it would be quite easy to convert to exponential form. Also, don’t forget that the values with get when we are done solving don’t always correspond to actual solutions so be careful!
Start Solution

First let’s notice that if we move the logarithm on the right side to the left side we can combine the two logarithms on the left side to get,

\[\begin{align*}\ln \left( {x - 1} \right) & = 1 + \ln \left( {3x + 2} \right)\\ \ln \left( {x - 1} \right) - \ln \left( {3x + 2} \right) & = 1\\ \ln \left( {\frac{{x - 1}}{{3x + 2}}} \right) & = 1\end{align*}\] Show Step 2

Now, we can easily convert this to exponential form (recall that because we are working with the natural logarithm the base is \(\bf{e}\)!).

\[\frac{{x - 1}}{{3x + 2}} = {{\bf{e}}^1} = {\bf{e}}\] Show Step 3

Now all we need to do is solve the equation from Step 2 and that is an equation that we know how to solve. Here is the solution work.

\[\begin{align*}\frac{{x - 1}}{{3x + 2}} & = {\bf{e}}\\ x - 1 & = {\bf{e}}\left( {3x + 2} \right)\\ x - 1 & = 3\,{\bf{e}}\,x + 2{\bf{e}}\\ x - 3\,{\bf{e}}\,x & = 1 + 2{\bf{e}}\\ \left( {1 - 3{\bf{e}}} \right)x & = 1 + 2{\bf{e}}\hspace{0.25in} \to \hspace{0.25in}x = \frac{{1 + 2{\bf{e}}}}{{1 - 3{\bf{e}}}} = - 0.89961\end{align*}\]

Do not get excited about the \(\bf{e}\) in the equation. It works the same as if it was just a 4 or 5 or any other number. The only real difference is that the answer is a little messier that we usually get with these kinds of problems. Also, for the next step it is probably best to convert these kinds of numbers into decimal form.

Show Step 4

As the final step we need to take the number from the above step and plug it into the original equation from the problem statement to make sure we don’t end up taking the logarithm of zero or negative numbers!

Here is the checking work for each of the numbers.

\(x = - 0.89961:\)

\[\begin{align*}\ln \left( { - 0.89961 - 1} \right) & = 1 + \ln \left( {3\left( { - 0.89961} \right) + 2} \right)\\ \ln \left( { - 1.89961} \right) & = 1 + \ln \left( { - 0.69883} \right)\hspace{0.25in}\,\,\,\,\,{\mbox{NOT OKAY}}\end{align*}\]

So, in this case the only number we got from Step 3 produced negative numbers in the logarithms and so can’t be a solution. What this means for us is that there is no solution to this equation. This happens on occasion and we shouldn’t worry about it when it does.

Note that it is vitally important that you do the check in the original equation. In the first step (where we combined two of the logarithms) we changed the equation and in the process introduced a number that is not in fact a solution.

Had we checked in any other equation in the solution work it would appear that \(x = - 0.89961\) would be a solution to the equation. However, that is only because we were checking in a “modified” equation and not the original equation which is what we were being asked to solve.

This is always the danger of modifying equations during the solution process. Unfortunately, with many logarithm equations that is our only solution path and so is something that we need to be prepared to deal with.