Section 6.4 : Solving Logarithm Equations
7. Solve the following equation.
\[\log \left( x \right) = 2 - \log \left( {x - 21} \right)\]Show All Steps Hide All Steps
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,
\[\begin{align*}\log \left( x \right) & = 2 - \log \left( {x - 21} \right)\\ \log \left( x \right) + \log \left( {x - 21} \right) & = 2\\ \log \left( {x\left( {x - 21} \right)} \right) & = 2\end{align*}\] Show Step 2Now, we can easily convert this to exponential form (recall that because there is no base given it is assumed to be 10!).
\[x\left( {x - 21} \right) = {10^2} = 100\] Show Step 3Now 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.
\[\begin{align*}x\left( {x - 21} \right) & = 100\\ {x^2} - 21x - 100 & = 0\\ \left( {x - 25} \right)\left( {x + 4} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = 25,\,\,\,\,x = - 4\end{align*}\] Show Step 4As 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.
\(x = 25:\)
\[\begin{align*}\log \left( {25} \right) & = 2 - \log \left( {25 - 21} \right)\\ \log \left( {25} \right) & = 2 - \log \left( 4 \right)\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}\]\(x = - 4:\)
\[\begin{align*}\log \left( { - 4} \right) & = 2 - \log \left( { - 4 - 21} \right)\\ \log \left( { - 4} \right) & = 2 - \log \left( { - 25} \right)\hspace{0.25in}\,\,\,\,\,{\mbox{NOT OKAY}}\end{align*}\]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 : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 25}}\).
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 \(x = - 4\) 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.