Solving a Logarithmic Equation or Inequality

A logarithmic equation or inequality contains a variable in the argument of a logarithm.

Since a logarithm’s argument is strictly greater than |0,| restrictions must be placed on the equation’s variable so the logarithm has a positive argument.

The restrictions must be set before solving the logarithmic equation.

1. Calculate the restrictions.

2. Reduce the expression using the laws of logarithms, if necessary.

3. Switch to exponential form.

4. Solve the equation.

5. Verify the solution(s).

6. Give the solution.

Solve the equation |\log_2(x+2)=4.|

1. Calculate the restrictions
   ||\begin{align}x+2&>0\\ x &> -2 \end{align}||

2. Reduce the expression
   The expression is already reduced.

3. Switch to exponential form
   ||\begin{align}\log_\color{#38B7CD}2(\color{#EC0000}{x+2})&=\color{#3A9A38}4\\ \Updownarrow \\\color{#EC0000}{x+2}&= \color{#38B7CD}2^\color{#3A9A38}4\end{align}||

4. Solve the equation
   ||\begin{align}x+2 &= 2^4\\ x+2 &= 16\\ x &= 14\end{align}||

5. Verify the solution
   The restriction is satisfied since |14>-2.|

6. Give the solution
   The solution is |x=14.|

There are two ways to verify the solution.

1. Ensure that the solution meets the restriction determined in step 1, as indicated in the previous example.

2. Substitute the solution into the original equation. In the previous example, the variable |x| is replaced in the original equation and the left-hand side is equivalent to the right-hand side.
   ||\begin{align}\log_2(\color{#3B87CD}x+2)&=4\\ \log_2(\color{#3B87CD}{14}+2)&\overset{?}{=}4\\ \log_2(16)&\overset{?}{=}4\\ 4&=4\end{align}||The solution is valid.

Solve the equation |\log_6 (x-1) + \log_6 (x) =1.|

1. Calculate the restrictions
   Since the equation has two logarithms containing the variable, calculate two restrictions. ||\begin{align}x-1&>0 \\ x&>1\\\\ x&>0\end{align}||Retain only |x>1| , because the restriction takes precedence over |x>0.|

2. Reduce the expression using the laws of logarithms
   The two logarithms have the same base, so use the product rule for logarithms. ||\begin{align}\log_6 (x-1) + \log_6 (x)&=1\\\ \log_6 \big((x-1)\times x\big)&=1\\ \log_6 (x^2-x)&=1 \end{align}||

3. Switch to exponential form
   ||\begin{align}\log_{\color{#3b87cd}6} (\color{#ec0000}{x^2-x})&=\color{#3a9a38}1\\ \Updownarrow\\ \color{#ec0000}{x^2-x}&=\color{#3b87cd}6^\color{#3a9a38}1\end{align}||

4. Solve the equation
   ||\begin{gather}x^2-x=6^1\\ x^2-x-6=0\\ (x-3)(x+2)=0\\ \overbrace{ \begin{aligned} x-3&=0\\x_1&=3 \end{aligned} \qquad \begin{aligned} x+2&=0\\x_2&=-2 \end{aligned} } \end{gather}||

5. Verify the solutions
   The restriction is respected for |x_1=3,| because |3>1.| However, it is not respected for |x_2=-2,| because |-2\not >1.| |x_2= -2| is rejected.

6. Give the solution
   The solution is |x=3.|

Solve the equation |\log(x-3)=\log(6x).|

1. Calculate the restrictions
   The equation has two logarithms containing the variable, so calculate two restrictions. ||\begin{align}x-3&>0\\x&>3\\\\ 6x&>0 \\ x&>0\end{align}||Retain only the restriction |x>3| for the rest of the solution, because this restriction takes precedence over |x>0.|

2. Reduce the expression using the laws of logarithms
   ||\begin{align}\log (x-3) &= \log (6x)\\ \log (x-3)-\log(6x)&=0\end{align}||The two logarithms have the same base, thus, use the quotient rule for logarithms. ||\begin{align}\log (x-3)-\log(6x)&=0\\\log \left(\dfrac{x-3}{6x}\right)&=0\end{align}||

3. Switch to exponential form
   Here, the base of the logarithm is not written. In this case, by convention, it is 10. ||\begin{align}\log_{\color{#3b87cd}{10}} \left(\color{#ec0000}{\dfrac{x-3}{6x}}\right)&=\color{#3a9a38}0\\ \Updownarrow\\ \color{#ec0000}{\dfrac{x-3}{6x}}&=\color{#3b87cd}{10}^\color{#3a9a38}0\end{align}||

4. Solve the equation
   ||\begin{align}\dfrac{x-3}{6x}&=10^0\\\dfrac{x-3}{6x}&=1\\x-3&=6x\\-3&=5x\\-0{.}6&=x \end{align}||

5. Verify the solution
   The restriction is not respected, because |-0{.}6\not >3.| Therefore, there is no solution.

6. Give the solution
   The equation has no solution.

Solve the equation |\log_4 (x^2+14x) = 1.|

1. Calculate the restrictions
   ||\begin{align} x^2+14x&>0\\ x(x+14)&>0\end{align}||Since the product of factors |x| and |(x+14)| is positive, these factors must have the same sign. This is the case when |x>0| (both terms are positive and the product is positive) and when |x<-14| (both terms are negative and thus the product is positive).

2. Reduce the expression
   The expression is already reduced.

3. Switch to exponential form
   ||\begin{align}\log_\color{#3b87cd}4 (\color{#ec0000}{x^2+14x})&=\color{#3a9a38}1\\ \Updownarrow\\\color{#ec0000}{x^2+14x} &= \color{#3b87cd}4^\color{#3a9a38}1\end{align}||

4. Solve the equation
   ||\begin{align}x&^2+14x=4^1\\ x&^2+14x-4=0\\\\ x&= \dfrac{-b\pm \sqrt{b^2- 4ac}}{2a}\\x&= \dfrac{-14\pm \sqrt{14^2- 4(1)(-4)}}{2(1)}\\ x&= \dfrac{-14\pm \sqrt{212}}{2}\\ x_1&\approx 0{.}28 \qquad x_2\approx -14{.}28\end{align}||

5. Verify the solution
   The restrictions are respected, because |0{.}28>0| and |-14{.}28<-14.|

6. Give the solution set
   The solution set is |x\in\{-14{.}28, 0{.}28\}.|

Logarithms with the same base are equal if the arguments have the same value. ||\log_c (a) = \log_c (b)\ \Leftrightarrow\ a=b||

Solve the equation |2\log_4 (x) - \log_4 (x+4) = \log_4 (x-2).|

1. Calculate the restrictions
   ||\begin{align}x&>0\\\\ x+4&>0\\x&>-4\\\\x-2&>0\\x&>2\end{align}||We only retain the restriction |x>2| for the rest of the solution, because this restriction takes precedence over |x>-4| and |x>0.|

2. Reduce the expression using the laws of logarithms
   For the first logarithm, use the power rule for logarithms. ||2\log_4 (x)=\log_4(x^2)|| The two logarithms to the left of the equality have the same base, thus, use the quotient rule for logarithms. ||\begin{align}\log_4 (x^2)-\log_4(x+4)&=\log_4(x-2)\\ \log_4 \left(\dfrac{x^2}{x+4}\right)&=\log_4(x-2)\end{align}||The two logarithms on each side of the equality have the same base, so the two arguments necessarily have the same value. This results in the following new equation. ||\dfrac{x^2}{x+4} = x-2||

3. Switch to exponential form
   Here, this step is unnecessary.

4. Solve the equation
   ||\begin{align}\dfrac{x^2}{x+4} &= x-2\\ x^2&=(x-2)(x+4)\\x^2&=x^2+2x-8\\0&=2x-8\\-2x&=-8\\x&=4\end{align}||

5. Verify the solution
   The restriction is satisfied, because |4>2.|

6. Give the solution
   The solution is |x=4.|

Solve the equation |2^{\log_2 5} + 5^{\log_5 3x} = 6^{\log_6 8}.|

1. Calculate the restrictions
   The equation has a single logarithm containing one variable, so there is only one restriction to calculate. ||\begin{align}3x&>0\\x&>0\end{align}||

2. Reduce the expression using the laws of logarithms
   According to the definition of logarithms, the expression |\log_2 5| refers to the exponent that must be applied to a base of |2| in order to obtain an answer of |5.| Since this expression is the exponent given to a base of |2,|, |2^{\log_2 5}=5.| In general, |c^{\log_c m} = m.| Applying this logarithmic identity to all the terms of the equation results in: ||\begin{align}2^{\log_2 5} + 5^{\log_5 3x}&=6^{\log_6 8}\\5 + 3x &= 8\end{align}||

3. Switch to exponential form
   Here, this step is unnecessary.

4. Solve the equation
   ||\begin{align}5 + 3x &= 8\\3x&=3\\x&=1\end{align}||

5. Verify the solution
   The restriction is satisfied because |1>0.|

6. Give the solution
   The solution is |x=1.|

1. Replace the inequality symbol with the equal symbol.

2. Calculate restrictions.

3. Reduce the expression using the laws of logarithms, if necessary.

4. Change to exponential form.

5. Solve the equation.

6. Verify the solution(s) of the equation.

7. Determine the solution set using a graph or number line.

Solve the inequality |\log_2 (x-2) \geq 4.|

1. Replace the inequality symbol with the equal symbol
   ||\log_2 (x-2) = 4||

2. Calculate the restrictions
   ||\begin{align}x-2 &>0\\ x&>2\end{align}||

3. Reduce the expression
   The expression is already reduced.

4. Switch to exponential form
   ||\begin{align}\log_\color{#3b87cd}2 (\color{#ec0000}{x-2}) &= \color{#3a9a38}4\\\color{#ec0000}{x-2}&=\color{#3b87cd}2^\color{#3a9a38}4\end{align}||

5. Solve the equation
   ||\begin{align}x-2&=2^4\\x-2&=16\\x&=18\end{align}||

6. Verify the solution
   The restriction is satisfied because |18>2.|

7. Determine the solution set
   Draw the graph of the logarithmic function to determine the solution set.

To determine the solution set (step 7), it is also possible to use a number line. Here is how the previous example can be solved using this method.

Draw a number line and place the restriction, determined in step 2, on it. Since |x>2,| |2| is excluded, so it is represented on the line with an empty point. Place the solution obtained in step 5 as well. Given that the initial inequality symbol is |\geq,| this solution is included , so use a solid point.

|2| and |18| are called critical values. They separate the line into three sections: |]-\infty, 2[,| |]2, 18]| , and |[18, +\infty[.|

For each section, replace |x| in the starting inequality with any value inside the interval. If the inequality is respected, the interval will be part of the solution set. If the inequality is not respected, the interval will not be part of the solution set.

• For interval |]-\infty, 2[,| replace |x| with |0.| ||\begin{align}\log_2 (\color{#3b87cd}x-2) \geq 4\\ \log_2 (\color{#3b87cd}0-2) \overset{?}{\geq} 4 \end{align}||The solution cannot be continued, since the argument of the logarithm is negative – which is impossible. Therefore, the interval |]-\infty, 2[| is not part of the solution set.

• For the interval |]2, 18],| replace |x| with |3.| This gives |0 \geq 4,| which is not possible. Thus, the interval |]2, 18]| is not part of the solution set.

• For the interval |[18, +\infty[,| replace |x| with |34.| This results in |5\geq\ 4,| which is true. So the interval |[18, +\infty[| is part of the solution set.

The solution set corresponds to |[18, +\infty[.|
Colour the section that corresponds to the solution set.

Solve the inequality |- \log_3 (2x-1) + 5 > 2.|

1. Replace the inequality symbol with the equal symbol
   ||- \log_3 (2x-1) + 5=2||

2. Calculate the restrictions
   ||\begin{align}2x-1&>0\\2x&>1\\x&>\dfrac{1}{2}\end{align}||

3. Reduce the expression
   ||\begin{align}-\log_3 (2x-1) + 5&=2\\-\log_3 (2x-1)&=-3\\\log_3 (2x-1)&=3\end{align}||

4. Switch to exponential form
   ||\begin{align}\log_\color{#3b87cd}3 (\color{#ec0000}{2x-1})&=\color{#3a9a38}3\\\Updownarrow\\\color{#ec0000}{2x-1}&=\color{#3b87cd}3^\color{#3a9a38}3\end{align}||

5. Solve the equation
   ||\begin{align}2x-1&=27\\2x&=28\\x&=14\end{align}||

6. Verify the solution
   The restriction is satisfied because |14>\dfrac{1}{2}.|

7. Determine the solution set