Now, we can easily convert this to exponential form.

\[ - x\left( {6 - x} \right) = {4^2} = 16\] Show Step 3

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

\[\begin{align*} - x\left( {6 - x} \right) & = 16\\ {x^2} - 6x - 16 & = 0\\ \left( {x - 8} \right)\left( {x + 2} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = 8,\,\,\,\,x = - 2\end{align*}\] Show Step 4

As the final step we need to take each of the numbers from the above step and plug them 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.

\(x = 8:\)

\[\begin{align*}{\log _4}\left( { - 8} \right) + {\log _4}\left( {6 - 8} \right) & = 2\\ {\log _4}\left( { - 8} \right) + {\log _4}\left( { - 2} \right) & = 2\hspace{0.25in}\,\,\,\,\,{\mbox{NOT OKAY}}\end{align*}\]

\(x = - 2:\)

\[\begin{align*}{\log _4}\left( { - \left( { - 2} \right)} \right) + {\log _4}\left( {6 - \left( { - 2} \right)} \right) & = 2\\ {\log _4}\left( 2 \right) + {\log _4}\left( 8 \right) & = 2\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}\]

In this case, the only one number did not produce negative numbers in the logarithms so that is the only number that will be a solution. The number that produced negative numbers in the logarithm is not a solution.

Therefore, the only solution to the equation is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 2}}\).

Be careful to not make the mistake of assuming that just because a value of \(x\) is negative that it will automatically not be a solution to the equation and just because a value of \(x\) is positive it will automatically be a solution. As we’ve shown here we can have negative values of \(x\) that are solutions and positive values of \(x\) that are not solutions.

Also 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 \(x = 8\) 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.