Solve Logarithmic Equations – Detailed Step-by-Step Solutions
This page presents worked examples on how to solve
logarithmic equations,
including several challenging problems. Examples 4, 5, 6, and 7 involve
logarithms with different bases and require the change-of-base formula.
Example 1:
Solve the logarithmic equation
\[
\log_{2}(x - 1) = 5
\]
Solution
-
Rewrite the logarithmic equation in exponential form:
\[
x - 1 = 2^{5}
\]
-
Solve for \(x\):
\[
x = 33
\]
-
Check:
\[
\log_{2}(33 - 1) = \log_{2}(32) = \log_{2}(2^{5}) = 5
\]
-
Conclusion: The solution is \(x = 33\).
Example 2:
Solve the logarithmic equation
\[
\log_{5}(x - 2) + \log_{5}(x + 2) = 1
\]
Solution
-
Apply the product rule:
\[
\log_{5}\big((x - 2)(x + 2)\big) = 1
\]
-
Rewrite in exponential form:
\[
(x - 2)(x + 2) = 5^{1}
\]
-
Simplify:
\[
x^{2} - 4 = 5 \quad \Rightarrow \quad x^{2} = 9
\]
-
Solve:
\[
x = 3 \quad \text{or} \quad x = -3
\]
-
Check:
-
For \(x = 3\):
\[
\log_{5}(1) + \log_{5}(5) = 0 + 1 = 1
\]
-
For \(x = -3\), the logarithms
\(\log_{5}(-5)\) and \(\log_{5}(-1)\) are undefined.
-
Conclusion: The only solution is \(x = 3\).
Example 3:
Solve the logarithmic equation
\[
\log_{3}(x - 2) + \log_{3}(x - 4)
= \log_{3}(2x^{2} + 139) - 1
\]
Solution
-
Rewrite \(1\) as \(\log_{3}(3)\):
\[
\log_{3}(x - 2) + \log_{3}(x - 4)
= \log_{3}(2x^{2} + 139) - \log_{3}(3)
\]
-
Apply product and quotient rules:
\[
\log_{3}\big((x - 2)(x - 4)\big)
= \log_{3}\!\left(\frac{2x^{2} + 139}{3}\right)
\]
-
Equate arguments:
\[
(x - 2)(x - 4) = \frac{2x^{2} + 139}{3}
\]
-
Multiply by 3 and simplify:
\[
3(x - 2)(x - 4) = 2x^{2} + 139
\]
-
Solving gives:
\[
x = -5 \quad \text{or} \quad x = 23
\]
-
Check:
- \(x = -5\) is invalid (negative logarithm arguments).
-
For \(x = 23\):
\[
\log_{3}(21 \cdot 19) = \log_{3}(399)
\]
and
\[
\log_{3}\!\left(\frac{1197}{3}\right) = \log_{3}(399)
\]
-
Conclusion: The solution is \(x = 23\).
Example 4:
Solve the logarithmic equation
\[
\log_{4}(x + 1) + \log_{16}(x + 1) = \log_{4}(8)
\]
Solution
-
Convert base 16 to base 4:
\[
\log_{16}(x + 1)
= \frac{\log_{4}(x + 1)}{\log_{4}(16)}
= \frac{1}{2}\log_{4}(x + 1)
\]
-
Rewrite the equation:
\[
\log_{4}(x + 1) + \frac{1}{2}\log_{4}(x + 1) = \log_{4}(8)
\]
-
Combine:
\[
\frac{3}{2}\log_{4}(x + 1) = \log_{4}(8)
\]
-
Convert to exponential form:
\[
(x + 1)^{3/2} = 8
\]
-
Solve:
\[
x = 3
\]
-
Conclusion: The solution is \(x = 3\).
Example 5:
Solve the logarithmic equation
\[
\log_{2}(x - 4) + \log_{\sqrt{2}}(x^{3} - 2)
+ \log_{0.5}(x - 4) = 20
\]
Solution
-
Use change of base:
\[
\log_{\sqrt{2}}(x^{3} - 2) = 2\log_{2}(x^{3} - 2)
\]
\[
\log_{0.5}(x - 4) = -\log_{2}(x - 4)
\]
-
Substitute and simplify:
\[
2\log_{2}(x^{3} - 2) = 20
\]
-
Solve:
\[
\log_{2}(x^{3} - 2) = 10
\]
\[
x^{3} - 2 = 2^{10}
\]
\[
x = \sqrt[3]{1026}
\]
Example 6:
Solve the logarithmic equation
\[
\ln(x + 6) + \log(x + 6) = 4
\]
Solution
-
Convert base 10 logarithm:
\[
\log(x + 6) = \frac{\ln(x + 6)}{\ln(10)}
\]
-
Solve:
\[
\ln(x + 6)\left(1 + \frac{1}{\ln(10)}\right) = 4
\]
\[
\ln(x + 6) = \frac{4\ln(10)}{1 + \ln(10)}
\]
-
Final answer:
\[
x = e^{\frac{4\ln(10)}{1 + \ln(10)}} - 6
\]
Example 7:
Solve the logarithmic equation
\[
\log_{5}(\ln(x + 3) - 1)
+ \log_{1/5}(\ln(x + 3) - 1) = 0
\]
Solution
-
Use change of base:
\[
\log_{1/5}(A) = -\log_{5}(A)
\]
-
The equation becomes:
\[
\log_{5}(A) - \log_{5}(A) = 0
\]
-
The equation is an identity whenever the arguments are defined:
\[
x > -3 \quad \text{and} \quad \ln(x + 3) > 1
\]
-
Solve:
\[
x > e - 3
\]
-
Conclusion:
The solution set is \((e - 3, +\infty)\).
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