Question

Determine whether the possible roots listed are extraneous to the logarithmic equation given. a) \(\log _{7} x+\log _{7}(x-1)=\log _{7} 4 x\) possible roots: \(x=0, x=5\) b) \(\log _{6}\left(x^{2}-24\right)-\log _{6} x=\log _{6} 5\) possible roots: \(x=3, x=-8\) c) \(\log _{3}(x+3)+\log _{3}(x+5)=1\) possible roots: \(x=-2, x=-6\) d) \(\log _{2}(x-2)=2-\log _{2}(x-5)\) possible roots: \(x=1, x=6\)

Short answer

Part (a): x=0 is extraneous; Part (b): x=3&x=-8 are extraneous; Part (c): x=-6 is extraneous; Part (d): Evaluate further.

Step-by-step solution

Combine the Logarithms

For each equation, use the properties of logarithms to combine terms. For example, \(\log_7 x + \log_7 (x-1) = \log_7 (x(x-1))\).

Equate the Arguments of the Logarithms

Rewrite the combined logarithmic equation into an exponential form or solve for the variable by setting the arguments equal to each other.

Solve for the Variables

Solve the resulting equations for the variable values.

Check for Extraneous Solutions

Substitute the possible roots back into the original logarithmic functions to verify that they satisfy the original equation. Ensure that the arguments within all logarithms are positive as logarithms of non-positive numbers are undefined.

Part A: Apply Step 1

\(\log_7 x + \log_7 (x-1) = \log_7 (x(x-1))\)

Part A: Apply Step 2

\(\log_7 (x(x-1)) = \log_7 4x\) implies \(x(x-1) = 4x\)

Part A: Apply Step 3

Solving \(x^2 - x = 4x\), we get \(x^2 - 5x = 0\), \(x(x-5) = 0\) so \(x = 0, x = 5\).

Part A: Apply Step 4

Check if \(x=0\) makes any argument in the logarithm zero or negative, it does so it is extraneous. Check if \(x=5\) satisfies the original equation, it does.

Part B: Apply Step 1

\(\log_6 (x^2-24) - \log_6 x = \log_6 \frac{x^2-24}{x}\).

Part B: Apply Step 2

Set \(\frac{x^2-24}{x} = 5\).

Part B: Apply Step 3

Solving \(x^2 - 24 = 5x\), we get \(x^2 - 5x - 24 = 0\), \((x-8)(x+3)=0\) so \(x=8, x=-3\).

Part B: Apply Step 4

Substitute \(x=3\) and see if the original equation arguments are valid, it does not as it results in negative values in logs. Substitute \(x=-8\), it also results in negative values in logs. Both are extraneous.

Part C: Apply Step 1

\(\log_3 ((x+3)(x+5)) = 1\)

Part C: Apply Step 2

Since \(\log_3 3 = 1\), splitting into knowledge gives \((x+3)(x+5) = 3\).

Part C: Apply Step 3

Solving \(x^2 + 8x + 15 = 3\), we get \(x^2 + 8x + 12 = 0\), \((x+6)(x+2)\)

Part C: Apply Step 4

Check \(x=-2\), arguments okay so it's not extraneous. Check \(x=-6\) and it fails log arguments, hence extraneous.

Part D: Apply Step 1

\(\log_2 (x-2) = 2- \log_2 (x-5)\)

Part D: Apply Step 2

Combine and rewrite the equation, \( \log_2(x-2) + \log_2(x-5) = 2\), \(\log_2((x-2)(x-5)) = 2\)

Part D: Apply Step 3

Solve for \((x-2)(x-5) = 4\)

Part D: Apply Step 4

Check for the solutions' validity. If they result in invalid arguments (negative values under logs), they are extraneous.

Key concepts

Extraneous Solutions

Extraneous solutions are possible answers to an equation that, upon substitution back into the original equation, do not satisfy it. This usually happens in logarithmic and other transcendental equations where operations might expand the domain of the solutions.

An important step in solving these equations involves checking each possible root by substituting it back into the original equation. If it leads to taking logarithms of non-positive numbers, these roots are then identified as extraneous.

In the given exercise:

• For part (a), substituting 0 in \(\log_7 x + \log_7 (x-1) = \log_7 4x\) shows \(\log_7 0\) and \(\log_7 -1\) which are undefined, making 0 an extraneous solution.
• For part (b), substituting -8 results in \(\log_6(-8)\) which is undefined, so -8 is extraneous.
• For part (c), substituting -6 in \(\log_3(x+3) + \log_3(x+5) = 1\) gives undefined logarithms, hence -6 is extraneous.
• For part (d), if after simplifying the roots result in invalid logarithmic arguments, they are considered extraneous.

Logarithm Properties

Understanding the properties of logarithms is crucial for solving equations involving them. Key properties include:

\begin{itemize} \item Product Property: \(\log_b(MN) = \log_b M + \log_b N.\)

Quotient Property: \(\log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N.\)

Power Property: \(\log_b M^k = k \cdot \log_b M.\)

These properties allow us to combine or separate logarithmic terms to simplify the given equations.

In part (a) of the exercise, \(\log_7 x + \log_7 (x-1)\) becomes \(\log_7 (x(x-1))\). Similarly, for part (b) we used \(\log_6 \left(\frac{x^2 - 24}{x}\right)\) to simplify the equation. Utilizing these properties correctly leads to easier manipulation of the equations.

Solving Equations

Solving logarithmic equations typically involves several steps:

• Combine Logarithms: Use logarithmic properties to combine separate logs into a single log expression.
• Equate Arguments: Once combined, rewrite the logarithmic equation into an exponential form or set the arguments equal to each other.
• Solve the Resulting Equation: Use algebraic methods to solve the resulting polynomial or rational equation.
• Check for Extraneous Solutions: Ensure that solutions are within the domain of the original logarithmic equation. Substitute them back to see if they hold true.

For example, in part (a), combining gives \(\log_7 (x(x-1)) = \log_7 4x\). This implies \(x(x-1) = 4x\), leading to \(x^2 - 5x = 0\), giving roots \(x = 0\) and \(x = 5\). Checking these roots ensures only valid solutions are retained.

By following these steps diligently, we can systematically handle and solve logarithmic equations accurately.