Solved Examples: Identifying Extraneous Solutions in Logarithmic Equations

📚 Quick Study Guide

• 🔑 Definition: An extraneous solution is a solution that emerges from the process of solving a problem but is not a valid solution to the original problem. In logarithmic equations, this often happens when the potential solution results in taking the logarithm of a negative number or zero, which is undefined.
• 📝 Checking Solutions: Always substitute potential solutions back into the original logarithmic equation to verify they are valid.
• ➕ Logarithmic Properties: Remember key properties like $\log_b{x} + \log_b{y} = \log_b{(xy)}$, $\log_b{x} - \log_b{y} = \log_b{(\frac{x}{y})}$, and $a\log_b{x} = \log_b{x^a}$. These are crucial for simplifying equations.
• 🚫 Invalid Arguments: Logarithms are only defined for positive arguments. If substituting a potential solution results in $\log(negative)$ or $\log(0)$, that solution is extraneous.
• 💡 Solving Steps:
  1. Combine logarithmic terms using properties of logarithms.
  2. Convert the logarithmic equation to an exponential equation.
  3. Solve the resulting algebraic equation.
  4. Check all potential solutions in the original equation.

Practice Quiz

1. What is an extraneous solution in the context of logarithmic equations?
   1. A) A solution that satisfies the original equation.
   2. B) A solution that makes the argument of a logarithm negative or zero.
   3. C) A solution that is always valid.
   4. D) A solution obtained by graphical methods only.
2. Solve for $x$: $\log_2(x) + \log_2(x-2) = 3$
   1. A) $x = -2$
   2. B) $x = 4$
   3. C) $x = 0, 4$
   4. D) $x = -2, 4$
3. Which of the following values of $x$ would be an extraneous solution for the equation $\log(x+3) + \log(x-3) = \log(7)$?
   1. A) $x = 4$
   2. B) $x = -4$
   3. C) $x = 2$
   4. D) $x = 3$
4. Solve for $x$: $\log_3(x+5) = 2$
   1. A) $x = 1$
   2. B) $x = 4$
   3. C) $x = -2$
   4. D) $x = -5$
5. Identify any extraneous solutions after solving the equation $\log(x) + \log(x-3) = 1$.
   1. A) $x = -2$
   2. B) $x = 5$
   3. C) $x = -2, 5$
   4. D) $x = 5$ is the only solution; there are no extraneous solutions.
6. Solve: $\log_4(x) + \log_4(x-6) = 2$
   1. A) $x = 8$
   2. B) $x = -2$
   3. C) $x = -2, 8$
   4. D) No solution
7. Determine the extraneous solution, if any, when solving $\log(2x+1) = 1 + \log(x-2)$.
   1. A) $x = 3$
   2. B) $x = -3$
   3. C) $x = \frac{1}{2}$
   4. D) There is no extraneous solution.