Solving Logarithmic Equations
We will now look at some equations that already have logarithms in them to start with. Thinking back to some of the restrictions we talked about with logarithms (the base and argument need to be positive numbers and the base cannot be \(1\)), we will need to think about extraneous solutions when we solve these.
The first type of equation we're going to look at is when both sides of the equation are in logarithmic form. When the logs have the same base and are equal to each other, then the argument of those logs must also be equal.
Example 1: Solve: \(\log_2\left(x^2+2x\right)=\log_2\left(3x+2\right)\).
\(\begin{align}x^2+2x&=3x+2\ & \ \ &\text{1) Since both logs have the same base and must be equal to each other, the argument}\\
& \ & \ \ &\text{of those logs must be equal.}\\
\hspace{4ex}x^2-x-2&=0 \ & \ \ &\text{2) Set equal to zero.}\\
\left(x-2\right)\left(x+1\right)&=0 \ & \ \ &\text{3) Factor.}\\
x&=2,-1\ & \ \ &\text{4) Solve, but remember that we have to check for extraneous solutions.}\end{align}\)
Check for Extraneous Solutions
\(x=2\)
\(\log_2\left(2^2+2\left(2\right)\right)=\log_2\left(3\left(2\right)+2\right)\)
\(\log_2\left(8\right)=\log_2\left(8\right)\)
This is a true statement and does not contradict any of the restrictions for a logarithm so \(x=2\) is a solution.
\(x=-1\)
\(\log_2\left(\left(-1\right)^2+2\left(-1\right)\right)=\log_2\left(3\left(-1\right)+2\right)\)
\(\log_2\left(-1\right)=\log_2\left(-1\right)\)
Even though the argument on each side is \(-1\), the argument cannot be a negative number. This is because the base, in this case \(2\), raised to any power will never be \(-1\). Therefore, \(x=-1\) is an extraneous solution.
The only solution to this equation is \(x=2\).
Example 2: Solve the equation: \(\ln\left(3x^2+9x-1\right)=\ln\left(5x+3\right)\).
We will now look at some equations that already have logarithms in them to start with. Thinking back to some of the restrictions we talked about with logarithms (the base and argument need to be positive numbers and the base cannot be \(1\)), we will need to think about extraneous solutions when we solve these.
The first type of equation we're going to look at is when both sides of the equation are in logarithmic form. When the logs have the same base and are equal to each other, then the argument of those logs must also be equal.
Example 1: Solve: \(\log_2\left(x^2+2x\right)=\log_2\left(3x+2\right)\).
\(\begin{align}x^2+2x&=3x+2\ & \ \ &\text{1) Since both logs have the same base and must be equal to each other, the argument}\\
& \ & \ \ &\text{of those logs must be equal.}\\
\hspace{4ex}x^2-x-2&=0 \ & \ \ &\text{2) Set equal to zero.}\\
\left(x-2\right)\left(x+1\right)&=0 \ & \ \ &\text{3) Factor.}\\
x&=2,-1\ & \ \ &\text{4) Solve, but remember that we have to check for extraneous solutions.}\end{align}\)
Check for Extraneous Solutions
\(x=2\)
\(\log_2\left(2^2+2\left(2\right)\right)=\log_2\left(3\left(2\right)+2\right)\)
\(\log_2\left(8\right)=\log_2\left(8\right)\)
This is a true statement and does not contradict any of the restrictions for a logarithm so \(x=2\) is a solution.
\(x=-1\)
\(\log_2\left(\left(-1\right)^2+2\left(-1\right)\right)=\log_2\left(3\left(-1\right)+2\right)\)
\(\log_2\left(-1\right)=\log_2\left(-1\right)\)
Even though the argument on each side is \(-1\), the argument cannot be a negative number. This is because the base, in this case \(2\), raised to any power will never be \(-1\). Therefore, \(x=-1\) is an extraneous solution.
The only solution to this equation is \(x=2\).
Example 2: Solve the equation: \(\ln\left(3x^2+9x-1\right)=\ln\left(5x+3\right)\).
The second type of equation is when only one side of the equation is written in logarithmic form. To solve, you can rewrite in exponential form and solve from there. Again, be careful to check for extraneous solutions.
Example 3: Solve \(\log_4\left(5x-3\right)=\frac{3}{2}\).
\(\begin{align}4^{\frac{3}{2}}&=5x-3\ & \ \ \ &\text{1) Rewrite in exponential form.}\\
8&=5x-3\ & \ \ \ &\text{2) Evaluate the exponent.}\\
5x&=11\ & \ \ \ &\text{3) Add.}\\
x&=\frac{11}{5}\ & \ \ \ &\text{4) Divide.}\end{align}\)
Check for Extraneous Solutions
\(x=\frac{11}{5}\)
\(\log_4\left(5\left(\frac{11}{5}\right)-3\right)=\frac{3}{2}\)
\(\log_4\left(8\right)=\frac{3}{2}\)
This is a true statement and does not contradict any of the restrictions so \(x=\frac{11}{5}\) is a solution.
The third type of equation we will look at is when there are multiple logarithms on one side and a non-zero constant on the other side of the equation. We must first use properties of logarithms to combine the multiple logarithms and then solve from there.
Example 4: Solve: \(\log x+\log\left(x+21\right)=2\).
\(\begin{align}&\log\left(x\left(x+21\right)\right)=2 \ & \ \ \ &\text{1) Combine using the Product Property for logarithms.}\\
&10^2=x^2+21x \ & \ \ \ &\text{2) Rewrite in exponential form…remember that the base of a common log is 10.}\\
&x^2+21x-100=0 \ & \ \ \ &\text{3) Set equation equal to zero.}\\
&\left(x+25\right)\left(x-4\right)=0 \ & \ \ \ &\text{4) Factor.}\\
&x=-25,4 \ & \ \ \ &\text{5) Solve.}\end{align}\)
Check for Extraneous Solutions
\(x=-25\)
\(\log\left(-25\right)+\log\left(-25+21\right)=2\)
The argument in a logarithm must be positive so \(x=-25\) is an extraneous solution.
\(x=4\)
\(\log\left(4\right)+\log\left(4+21\right)=2\)
This does not contradict any of the restrictions for a logarithm (both arguments are positive numbers) so \(x=4\) is a solution.
Example 5: Solve the equation: \(\log_3\left(5x-4\right)-\log_3\left(x+1\right)=2\).