Section 6.4 : Solving Logarithm Equations
9. Solve the following equation.
\[2\log \left( x \right) - \log \left( {7x - 1} \right) = 0\]Show All Steps Hide All Steps
First let’s notice that we can move the 2 in front of the first logarithm into the logarithm as follows,
\[\log \left( {{x^2}} \right) - \log \left( {7x - 1} \right) = 0\]We can now combine the two logarithms to get,
\[\log \left( {\frac{{{x^2}}}{{7x - 1}}} \right) = 0\] 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!).
\[\frac{{{x^2}}}{{7x - 1}} = {10^0} = 1\] 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.
\[\begin{align*}\frac{{{x^2}}}{{7x - 1}} & = 1\\ {x^2} & = 7x - 1\\ {x^2} - 7x + 1 & = 0\end{align*}\]We can’t factor this but we can use the quadratic formula on it. Doing that gives the following two numbers.
\[x = \frac{{7 \pm \sqrt {{7^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}} = \frac{{7 \pm \sqrt {45} }}{2}\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,x = 0.1459,\,\,\,\,\,\,\,x = 6.8541\]Don’t worry about the fact that we needed to use the quadratic formula to solve this. This will happen on occasion and we need to be able to deal with it when it happens.
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.
\(x = 0.1459:\)
\[\begin{align*}2\log \left( {0.1459} \right) - \log \left( {7\left( {0.1459} \right) - 1} \right) & = 0\\ 2\log \left( {0.1459} \right) - \log \left( {0.0213} \right) & = 0\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}\]\(x = 6.8541:\)
\[\begin{align*}2\log \left( {6.8541} \right) - \log \left( {7\left( {6.8541} \right) - 1} \right) & = 0\\ 2\log \left( {6.8541} \right) - \log \left( {46.9787} \right) & = 0\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}\]In this case, both numbers do not produce negative numbers in the logarithms and so they are in fact both solutions (won’t happen with every problem so don’t always expect this to happen!).
Therefore, the solutions to the equation are then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0.1459}}\) and \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 6.8541}}\).