# Logarithmic Equations

### Solving Simple Logarithmic Equations A simple logarithmic equation can be solved by using an equality property of logarithms or by writing the equation in exponential form.

A logarithmic equation is an equation with a logarithmic expression that contains a variable. A logarithmic equation may have logarithms on one side, such as $\log_{3}x=7$, or on both sides, such as $\log_{2}(x+1)=\log_{2} 6$.

Solving a logarithmic equation with logarithms on both sides may involve the equality property: For $x>0$ and $y>0$, if $\log_{b} x=\log_{b} y$, then $x=y$.

The argument of a logarithm is the expression the logarithm applies to. For example, the argument of $\log_4(x-3)$ is $x-3$. When solving equations involving logarithms, check the answers in the original equation to make sure the argument of the logarithm is positive.
Step-By-Step Example
Solve a Simple Equation with Logarithms on Both Sides
Solve the logarithmic equation:
$\log_{2}(x+1)=\log_{2}6$
Step 1

Use the equality property.

The logarithms have the same base, so the arguments are equal.
\begin{aligned}\log_{2}(x+1)&=\log_{2} 6\\x+1&=6\end{aligned}
Step 2
Solve the resulting equation.
$x=5$
Solution
Substitute the value of $x$ in the original equation to make sure all arguments are positive.
\begin{aligned}\log_{2}(5+1)&=\log_{2}6\\\log_{2}6&=\log_{2}6\end{aligned}
The arguments are positive. So, $x=5$ is a valid solution.
To solve a logarithmic equation with logarithms on one side, use the relationship between exponents and logarithms: If $\log_{b} x=\log_{b} y$, then $b^y=x$.
Step-By-Step Example
Solve a Simple Equation with a Logarithm on One Side
Solve the logarithmic equation:
$\log_{3}x=7$
Step 1
Write the equation in exponential form.
\begin{aligned}\log_{b}y&=x &\rightarrow &&b^{x}&=y\\ \log_{3}x&=7&\rightarrow &&3^{7}&=x\end{aligned}
Step 2
Simplify to determine the value of $x$.
\begin{aligned}3^{7}&=x\\2\text{,}187&=x\end{aligned}
Solution
Make sure the argument is positive by substituting the value of $x$ in the original equation.
\begin{aligned}\log_{3}x&=7\\\log_{3}2\text{,}187&=7\end{aligned}
The argument is positive. So, $x=2\text{,}187$ is a valid solution.

### Solving Multistep Logarithmic Equations A multistep logarithmic equation can be solved by using properties of logarithms and then using an equality property or writing in exponential form.
To solve multistep logarithmic equations, first use properties of logarithms to simplify the equation.

### Properties of Logarithms

Name Property
Product rule
$\log_{b}{xy}=\log_{b}{x}+\log_{b}{y}$
Quotient rule
$\log_{b}{\frac{x}{y}}=\log_{b}{x}-\log_{b}{y}$
Power rule
$\log_{b}{x^p}=p\cdot\log_{b}{x}$
Change of base rule
$\log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}$

After simplifying the equation, if there is a single logarithm on both sides of the equation, use the equality property: If $\log_{b} x=\log_{b} y$, then $x=y$. If there is a single logarithm on only one side of the equation, write the equation in exponential form: If $\log_{b} x= y$, then $b^{y}=x$. Make sure to check all answers to ensure that the argument of the logarithm is positive.
Step-By-Step Example
Solve a Multistep Equation with Logarithms on Both Sides
Solve the logarithmic equation:
$\log_{4}8x-\log_{4}(x-2)=\log_{4}x$
Step 1

Apply a property of logarithms.

The equation involves a difference of logarithms, and all logarithms have the same base. Use the quotient rule.
\begin{aligned}\log_{4}8x-\log_{4}(x-2)&=\log_{4}x\\\log_{4}{\left ( \frac{8x}{x-2} \right )}&=\log_{4}{x}\end{aligned}
Step 2

Use the equality property.

The logarithms have the same base, so the arguments are equal.
$\frac{8x}{x-2}=x$
Step 3
Multiply both sides by $x-2$.
\begin{aligned}\frac{8x}{x-2}&=x\\\left(\frac{8x}{x-2}\right)(x-2)&=x(x-2)\\8x&=x(x-2) \end{aligned}
Step 4
Distribute the $x$ on the right side of the equation.
\begin{aligned}8x&=x(x-2) \\8x&=x^{2}-2x\end{aligned}
Step 5
Set the equation equal to zero by subtracting $8x$ from both sides of the equation.
\begin{aligned}8x=x^{2}-\phantom{1}2x\\\underline{-8x=\phantom{x^2}-\phantom{1}8x}\\\phantom{-8}0=x^{2}-10x\end{aligned}
Step 6
Factor the equation.
\begin{aligned}0&=x^{2}-10x\\0&=x(x-10)\end{aligned}
Step 7
Use the zero product property to set each factor equal to zero. Then, solve each equation.
\begin{aligned}x=0\quad \text{ or }\quad x-10&=0\\x&=10\end{aligned}
Solution

Substitute the values of $x$ in the original equation to make sure all arguments are positive.

Check $x=0$:
$\log_{4}(8\cdot 0)-\log_{4}(0-2)=\log_{4}0$
The solution $x=0$ is not valid because $\log_{4} 0$ is undefined. Check $x=10$:
\begin{aligned}\log_{4}(8\cdot 10)-\log_{4}(10-2)&=\log_{4}10&&\text{Substitute }10 \text{ for }x\text{.}\\\log_{4}80-\log_{4}8&=\log_{4}10&&\text{Simplify.}\\\log_{4}\frac{80}{8}&=\log_{4}10&&\text{Use the quotient rule of logarithms.}\end{aligned}
The argument is positive. So, $x=10$ is a valid solution. It is the only valid solution for the equation.
Step-By-Step Example
Solve a Multistep Equation with Logarithms on One Side
Solve the logarithm:
$\log(x) + \log(x-15)=2$
Step 1

Apply a property of logarithms.

The equation involves a sum of logarithms. No base is shown, so all the logarithms in the equation have a base of 10. Use the product rule.
\begin{aligned}\log(x) + \log(x-15)&=2\\\log[(x)(x-15)]&=2\end{aligned}
Simplify.
$\log(x^2-15x)=2$
Step 2
Write the equation in exponential form.
$10^2=x^2-15x$
Step 3
Simplify.
\begin{aligned}&\begin{aligned}100&=x^2-15x &&\text{Simplify.}\\0&=x^2-15x-100 &&\text{Set the equation equal to zero.}\\0&=(x+5)(x-20)&&\text{Factor.}\end{aligned}\end{aligned}
Use the zero product property to set each factor equal to zero. Then, solve each equation.
\begin{aligned} x+5&=0\\x&=-5\end{aligned} \hspace{10pt} \text{or} \hspace{10pt} \begin{aligned}x-20&=0\\x&=20\end{aligned}
Solution

Substitute the values of $x$ in the original equation to make sure all arguments are positive.

Check $x=-5$:
$\log(-5) + \log(-5-15)=2$
The solution $x=-5$ is not valid because the arguments of $\log(-5)$ and $\log(-5-15)$ are negative. Check $x=20$:
\begin{aligned}\log(20) + \log(20-15)&=2\\\log(20)+\log(5)&=2\\\log(20 \cdot 5)&=2\\\log100&=2\end{aligned}
The solution $x=20$ is valid. It is the only valid solution for the equation.

### Solving Logarithmic Equations Using Substitution Some logarithmic equations can be solved by using substitution to change the form of the equation.
Some logarithmic equations can be solved by replacing a logarithmic expression with a temporary variable. Solve for the temporary variable, and then replace the temporary variable with the logarithmic expression and solve for $x$.
Step-By-Step Example
Solve a Logarithmic Equation Using Substitution
Solve the logarithmic equation:
$\ln^2(x)+3\ln(x)=-2$
Step 1

Use a temporary variable to represent the logarithmic expression.

Substitute $u$ for $\ln{x}$ in the original equation.
$u^2 + 3u =-2$
Step 2
Simplify.
\begin{aligned}u^2+&3u+2=0\;\;\;\;\;&&\text{Set the equation equal to zero.}\\\;(u + 2)(&u+1)= 0\;\;\;\;&&\text{Factor.}\end{aligned}
Use the zero product property to set each factor equal to zero and then solve each equation.
\begin{aligned}u+2&=0\\u&=-2\end{aligned}\hspace{10pt}\text{or}\hspace{10pt}\begin{aligned} u+1&=0\\u&=-1\end{aligned}
Step 3

Replace $u$ with the expression it represents.

Substitute $\ln{x}$ for $u$ in the solutions.
$\ln(x) = -2 \hspace{10pt}\text{or}\hspace{10pt}\ln(x) = -1$
Step 4
Write the equations in exponential form.
$x = e^{-2}\hspace{10pt} \text{or} \hspace{10pt} x = e^{-1}$
Write the exponential expressions with positive exponents.
$x = \frac{1}{e^2} \hspace{10pt}\text{or} \hspace{10pt}x = \frac{1}{e^1}$
Step 5

Substitute the values of $x$ in the original equation to make sure all arguments are positive.

Since $e$ is positive, $\frac{1}{e^2}$ and $\frac{1}{e}$ must be positive.

Check $x = \frac{1}{e^2}$:
\begin{aligned}\ln^2\left(\frac{1}{e^2}\right)+3\ln\left(\frac{1}{e^2}\right)&=-2\\4+(-6)&=-2\end{aligned}
The solution $x = \frac{1}{e^2}$ is valid. Check $x = \frac{1}{e}$:
\begin{aligned}\ln^2\left(\frac{1}{e}\right)+3\ln\left(\frac{1}{e}\right)&=-2\\1+(-3)&=-2\end{aligned}
The solution $x = \frac{1}{e}$ is also valid.
Solution
The solutions $x = \frac{1}{e^2}$ and $x = \frac{1}{e}$ are valid. The approximate solutions are $x \approx0.14$ and $x \approx0.37$.

### Solving Logarithmic Equations by Graphing Logarithmic equations can be solved by graphing the related functions for both sides of the equation and looking for points of intersection.
To solve a logarithmic equation by graphing, graph the related functions for both sides of the equation and look for points of intersection. This process is similar to solving a system of equations in two variables. If the logarithmic equation is set equal to zero, graph the related function for the logarithmic expression and look for the zeros. A zero of a function is any input value of a function that makes the output of the function equal zero.
Step-By-Step Example
Solve a Logarithmic Equation by Graphing
Solve for $x$:
$\ln{(x+5)}=1$
Step 1
Use a graphing calculator or other graphing utility to graph the functions:
$f(x)=\ln{(x+5)} \hspace{20pt} f(x)=1$
Step 2
Use the trace function on a graphing utility to approximate the $x$-value of the point of intersection.
Solution
The graphs intersect where $x \approx -2.28$.