How to Solve Logarithmic Equations — Detailed Step-by-Step Solutions
This page presents detailed worked examples on how to solve logarithmic equations, from foundational problems to challenging ones involving change of base.
Key rule: Always check your final answers. Logarithms are only defined for strictly positive arguments, so some algebraic solutions may be extraneous. Examples 4 through 7 use logarithms with different bases and demonstrate the change-of-base formula.
Example 1: Solve a Basic Logarithmic Equation
Solve the logarithmic equation:
$$ \log_{2}(x - 1) = 5 $$View Solution
- Rewrite the logarithmic equation in its exponential form: $$ x - 1 = 2^{5} $$
- Calculate the exponent and solve for $x$: $$ x - 1 = 32 $$ $$ x = 33 $$
Substitute $x = 33$ back into the original equation:
$$ \log_{2}(33 - 1) = \log_{2}(32) = \log_{2}(2^{5}) = 5 $$The solution is valid. Conclusion: $x = 33$
Example 2: Product Rule and an Extraneous Solution
Solve the logarithmic equation:
$$ \log_{5}(x - 2) + \log_{5}(x + 2) = 1 $$View Solution
- Apply the logarithmic product rule ($\log_b(m) + \log_b(n) = \log_b(m \cdot n)$): $$ \log_{5}\big((x - 2)(x + 2)\big) = 1 $$
- Rewrite in exponential form: $$ (x - 2)(x + 2) = 5^{1} $$
- Expand the left side (difference of squares) and solve for $x$: $$ x^{2} - 4 = 5 $$ $$ x^{2} = 9 $$ $$ x = 3 \quad \text{or} \quad x = -3 $$
- For $x = 3$: $\log_{5}(1) + \log_{5}(5) = 0 + 1 = 1$. (Valid)
- For $x = -3$: The logarithm becomes $\log_{5}(-5)$, which is undefined because you cannot take the logarithm of a negative number.
Conclusion: The only valid solution is $x = 3$.
Example 3: Combining Product and Quotient Rules
Solve the logarithmic equation:
$$ \log_{3}(x - 2) + \log_{3}(x - 4) = \log_{3}(2x^{2} + 139) - 1 $$View Solution
- Rewrite $1$ as $\log_{3}(3)$ so all terms have the same base: $$ \log_{3}(x - 2) + \log_{3}(x - 4) = \log_{3}(2x^{2} + 139) - \log_{3}(3) $$
- Apply the product rule on the left and the quotient rule on the right: $$ \log_{3}\big((x - 2)(x - 4)\big) = \log_{3}\!\left(\dfrac{2x^{2} + 139}{3}\right) $$
- Since the bases are identical, equate the arguments: $$ (x - 2)(x - 4) = \dfrac{2x^{2} + 139}{3} $$
- Multiply both sides by 3 to clear the denominator, then expand: $$ 3(x^{2} - 6x + 8) = 2x^{2} + 139 $$ $$ 3x^{2} - 18x + 24 = 2x^{2} + 139 $$
- Rearrange into standard quadratic form ($ax^2 + bx + c = 0$): $$ x^{2} - 18x - 115 = 0 $$
- Factor the quadratic equation: $$ (x - 23)(x + 5) = 0 $$ $$ x = 23 \quad \text{or} \quad x = -5 $$
- If $x = -5$, the term $\log_{3}(-5 - 2) = \log_{3}(-7)$ is undefined. This is an extraneous solution.
- If $x = 23$, all logarithmic arguments are positive: $\log_{3}(21) + \log_{3}(19) = \log_{3}(399)$ and $\log_{3}(1197/3) = \log_{3}(399)$.
Conclusion: The only solution is $x = 23$.
Example 4: Change of Base — Converting Base 16 to Base 4
Solve the logarithmic equation:
$$ \log_{4}(x + 1) + \log_{16}(x + 1) = \log_{4}(8) $$View Solution
- Use the change-of-base formula to convert base 16 to base 4: $$ \log_{16}(x + 1) = \dfrac{\log_{4}(x + 1)}{\log_{4}(16)} = \dfrac{\log_{4}(x + 1)}{2} = \dfrac{1}{2}\log_{4}(x + 1) $$
- Rewrite the original equation substituting this new form: $$ \log_{4}(x + 1) + \dfrac{1}{2}\log_{4}(x + 1) = \log_{4}(8) $$
- Combine like terms ($1 + \dfrac{1}{2} = \dfrac{3}{2}$): $$ \dfrac{3}{2}\log_{4}(x + 1) = \log_{4}(8) $$
- Use the power rule ($\dfrac{3}{2} \log_4(A) = \log_4(A^{3/2})$): $$ \log_{4}\big((x + 1)^{3/2}\big) = \log_{4}(8) $$
- Equate the arguments and solve: $$ (x + 1)^{3/2} = 8 $$ $$ (x + 1)^{3} = 8^{2} = 64 $$ $$ x + 1 = \sqrt[3]{64} = 4 $$ $$ x = 3 $$
$x + 1 = 4 > 0$, so both logarithms are defined. Conclusion: $x = 3$.
Example 5: Multiple Bases — √2 and 0.5
Solve the logarithmic equation:
$$ \log_{2}(x - 4) + \log_{\sqrt{2}}(x^{3} - 2) + \log_{0.5}(x - 4) = 20 $$View Solution
- Use the change-of-base formula to express all logarithms in base 2: $$ \log_{\sqrt{2}}(x^{3} - 2) = \dfrac{\log_{2}(x^{3} - 2)}{\log_{2}(2^{1/2})} = \dfrac{\log_{2}(x^{3} - 2)}{1/2} = 2\log_{2}(x^{3} - 2) $$ $$ \log_{0.5}(x - 4) = \dfrac{\log_{2}(x - 4)}{\log_{2}(2^{-1})} = -\log_{2}(x - 4) $$
- Substitute these conversions back into the equation: $$ \log_{2}(x - 4) + 2\log_{2}(x^{3} - 2) - \log_{2}(x - 4) = 20 $$
- Notice that $\log_{2}(x - 4)$ and $-\log_{2}(x - 4)$ cancel each other out: $$ 2\log_{2}(x^{3} - 2) = 20 $$
- Divide by 2: $$ \log_{2}(x^{3} - 2) = 10 $$
- Convert to exponential form and solve: $$ x^{3} - 2 = 2^{10} $$ $$ x^{3} - 2 = 1024 $$ $$ x^{3} = 1026 $$ $$ x = \sqrt[3]{1026} \approx 10.09 $$
Since $\sqrt[3]{1026} \approx 10.09$, we have $x - 4 \approx 6.09 > 0$ and $x^{3} - 2 = 1024 > 0$. Both arguments are positive, so the solution is valid.
Conclusion: $x = \sqrt[3]{1026}$.
Example 6: Mixing Natural Log and Common Log
Solve the logarithmic equation:
$$ \ln(x + 6) + \log(x + 6) = 4 $$View Solution
- Convert the common base-10 logarithm to the natural logarithm ($\ln$, base $e$): $$ \log(x + 6) = \dfrac{\ln(x + 6)}{\ln(10)} $$
- Substitute into the original equation: $$ \ln(x + 6) + \dfrac{\ln(x + 6)}{\ln(10)} = 4 $$
- Factor out $\ln(x + 6)$: $$ \ln(x + 6)\left(1 + \dfrac{1}{\ln(10)}\right) = 4 $$
- Isolate $\ln(x + 6)$ by getting a common denominator for the constant: $$ \ln(x + 6)\left(\dfrac{\ln(10) + 1}{\ln(10)}\right) = 4 $$ $$ \ln(x + 6) = \dfrac{4\ln(10)}{1 + \ln(10)} $$
- Convert back to exponential form to solve for $x$: $$ x + 6 = e^{\dfrac{4\ln(10)}{1 + \ln(10)}} $$ $$ x = e^{\dfrac{4\ln(10)}{1 + \ln(10)}} - 6 \approx 10.27 $$
$x + 6 \approx 16.27 > 0$, so both logarithms are defined.
Conclusion: $x = e^{\frac{4\ln 10}{1+\ln 10}} - 6$.
Example 7: Identity Equation with a Domain Restriction
Solve the logarithmic equation:
$$ \log_{5}(\ln(x + 3) - 1) + \log_{1/5}(\ln(x + 3) - 1) = 0 $$View Solution
- Use the change-of-base formula. A base of $1/5$ is $5^{-1}$, which makes the logarithm negative: $$ \log_{1/5}(A) = \dfrac{\log_{5}(A)}{\log_{5}(5^{-1})} = -\log_{5}(A) $$
- The equation simplifies drastically: $$ \log_{5}(\ln(x + 3) - 1) - \log_{5}(\ln(x + 3) - 1) = 0 $$
- This results in $0 = 0$, meaning the equation is an identity. It is true for all values of $x$ as long as the arguments are mathematically defined.
The argument inside the base-5 logarithm must be strictly greater than 0:
$$ \ln(x + 3) - 1 > 0 $$ $$ \ln(x + 3) > 1 $$Convert to an exponential inequality:
$$ x + 3 > e^{1} $$ $$ x > e - 3 $$Conclusion: The solution set is all real numbers in the interval $(e - 3, +\infty)$.
Frequently Asked Questions About Logarithmic Equations
What is the first step to solve a logarithmic equation?
Isolate a single logarithm on one side when possible, then rewrite the equation in exponential form. If several logarithms appear, combine them first using the product, quotient, or power rules.
Why do logarithmic equations sometimes have extraneous solutions?
Combining or rewriting logarithms can extend the domain of the equation. A value that solves the transformed equation may make an original argument zero or negative, which is undefined. Always substitute each candidate back into the original equation.
How do you use the change of base formula?
The change-of-base formula is $\log_b(A) = \dfrac{\log_c(A)}{\log_c(b)}$ for any positive base $c \neq 1$. Choose a base $c$ that matches the other terms in the equation so you can combine the logarithms.
What is the domain of a logarithmic equation?
Every argument of a logarithm must be strictly positive, and every base must be positive and not equal to 1. The solution set of a logarithmic equation is the intersection of all these domain restrictions.