Section 6.4 : Solving Logarithm Equations
2. Solve the following equation.
\[\log \left( {6x} \right) - \log \left( {4 - x} \right) = \log \left( 3 \right)\]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 difference of logarithms (provide the coefficient of each is a one of course…) as follows,
\[\log \left( {\frac{{6x}}{{4 - x}}} \right) = \log \left( 3 \right)\]We now have only two logarithms and each logarithm is on opposite sides of the equal sign and each has the same base, 10 in this case. Therefore, we can use this property to just set the arguments of each equal. Doing this gives,
\[\frac{{6x}}{{4 - x}} = 3\] Show Step 2Now all we need to do is solve the equation from Step 1 and that is an equation that we know how to solve. Here is the solution work.
\[\begin{align*}\frac{{6x}}{{4 - x}} & = 3\\ 6x & = 3\left( {4 - x} \right) = 12 - 3x\\ 9x & = 12\hspace{0.25in} \to \hspace{0.25in}x = \frac{{12}}{9} = \frac{4}{3}\end{align*}\] Show Step 3As 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 the number.
\(x = \frac{4}{3}:\)
\[\begin{align*}\log \left( {6\left( {\frac{4}{3}} \right)} \right) - \log \left( {4 - \frac{4}{3}} \right) & = \log \left( 3 \right)\\ \log \left( 8 \right) - \log \left( {\frac{8}{3}} \right) & = \log \left( 3 \right)\hspace{0.25in}{\mbox{OKAY}}\end{align*}\]In this case, the number did not produce negative numbers in the logarithms so it is in fact a solution (won’t happen with every problem so don’t always expect this to happen!).
Therefore, the solution to the equation is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{4}{3}}}\).