Section 6.4 : Solving Logarithm Equations
3. Solve the following equation.
\[\ln \left( x \right) + \ln \left( {x + 3} \right) = \ln \left( {20 - 5x} \right)\]Show All Steps Hide All Steps
Recall the property that says if \({\log _b}x = {\log _b}y\) then \(x = y\). That doesn’t appear to have any use here since there are three logarithms in the equation. However, recall that we can combine a sum of logarithms (provide the coefficient of each is a one of course…) as follows,
\[\ln \left( {x\left( {x + 3} \right)} \right) = \ln \left( {20 - 5x} \right)\]We now have only two logarithms and each logarithm is on opposite sides of the equal sign and each has the same base, \(\bf{e}\) in this case. Therefore, we can use this property to just set the arguments of each equal. Doing this gives,
\[x\left( {x + 3} \right) = 20 - 5x\] Show Step 2Now all we need to do is solve the equation from Step 1 and that is a quadratic equation that we know how to solve. Here is the solution work.
\[\begin{align*}x\left( {x + 3} \right) & = 20 - 5x\\ {x^2} + 3x & = 20 - 5x\\ {x^2} + 8x - 20 & = 0\\ \left( {x + 10} \right)\left( {x - 2} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = - 10,\,\,\,\,\,x = 2\end{align*}\] Show Step 3As 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.
\(x = - 10:\)
\[\begin{align*}\ln \left( { - 10} \right) + \ln \left( { - 10 + 3} \right) & = \ln \left( {20 - 5\left( { - 10} \right)} \right)\\ \ln \left( { - 10} \right) + \ln \left( { - 7} \right) & = \ln \left( {70} \right)\hspace{0.25in}\,\,\,\,\,{\mbox{NOT}}\,\,{\mbox{OKAY}}\end{align*}\]\(x = 2:\)
\[\begin{align*}\ln \left( 2 \right) + \ln \left( {2 + 3} \right) & = \ln \left( {20 - 5\left( 2 \right)} \right)\\ \ln \left( 2 \right) + \ln \left( 5 \right) & = \ln \left( {10} \right)\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}\]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 : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 2}}\).
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 \(x = - 10\) 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.