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### Section 6.4 : Solving Logarithm Equations

4. Solve the following equation.

${\log _3}\left( {25 - {x^2}} \right) = 2$

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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 and yes we can use that property here! 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 $$x = y$$. That doesn’t appear to have any use here since there is only one logarithm in the equation. Note however that we could write the right side, i.e. the 2, as,

$2 = {\log _3}\left( {{3^2}} \right)$

Doing this means we can write the equation as,

${\log _3}\left( {25 - {x^2}} \right) = {\log _3}\left( {{3^2}} \right) = {\log _3}\left( 9 \right)$

We now have two logarithms and each logarithm is on opposite sides of the equal sign and each has the same base, 3 in this case. Therefore, we can use this property to just set the arguments of each equal. Doing this gives,

$25 - {x^2} = 9$ 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*}25 - {x^2} & = 9\\ 16 & = {x^2}\hspace{0.25in} \to \hspace{0.25in}x = \pm \sqrt {16} = \pm 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.

$$x = - 4:$$

\begin{align*}{\log _3}\left( {25 - {{\left( { - 4} \right)}^2}} \right) & = 2\\ {\log _3}\left( 9 \right) & = 2\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}

$$x = 4:$$

\begin{align*}{\log _3}\left( {25 - {{\left( 4 \right)}^2}} \right) & = 2\\ {\log _3}\left( 9 \right) & = 2\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 : $$\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 4}}$$ and $$\require{bbox} \bbox[2pt,border:1px solid black]{{x = 4}}$$.

Be careful to not make the mistake of assuming that just because a value of $$x$$ is negative that it will automatically not be a solution to the equation. As we’ve shown here, even though $$x = - 4$$ is negative it did not produce any negative values in the logarithms and so is perfectly acceptable as a solution.