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Section 6.4 : Solving Logarithm Equations

1. Solve the following equation.

\[{\log _4}\left( {{x^2} - 2x} \right) = {\log _4}\left( {5x - 12} \right)\]

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Hint : We had a very nice property from the notes on how to solve equations that contained exactly two logarithms with the same base! Also, don’t forget that the values with get when we are done solving logarithm equations don’t always correspond to actual solutions to the equation so be careful!
Start Solution

Recall the property that says if \({\log _b}x = {\log _b}y\) then \(x = y\). Since each logarithm is on opposite sides of the equal sign and each has the same base, 4 in this case, we can use this property to just set the arguments of each equal. Doing this gives,

\[{x^2} - 2x = 5x - 12\] Show Step 2

Now 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.

\[\begin{align*}{x^2} - 2x & = 5x - 12\\ {x^2} - 7x + 12 & = 0\\ \left( {x - 3} \right)\left( {x - 4} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = 3,\,\,\,\,x = 4\end{align*}\] Show Step 3

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.

\(x = 3:\)

\[\begin{align*}{\log _4}\left( {{{\left( 3 \right)}^2} - 2\left( 3 \right)} \right) & = {\log _4}\left( {5\left( 3 \right) - 12} \right)\\ {\log _4}\left( 3 \right) & = {\log _4}\left( 3 \right)\hspace{0.25in}{\mbox{OKAY}}\end{align*}\]

\(x = 4:\)

\[\begin{align*}{\log _4}\left( {{{\left( 4 \right)}^2} - 2\left( 4 \right)} \right) & = {\log _4}\left( {5\left( 4 \right) - 12} \right)\\ {\log _4}\left( 8 \right) & = {\log _4}\left( 8 \right)\hspace{0.25in}{\mbox{OKAY}}\end{align*}\]

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 : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 3}}\) and \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 4}}\).