Question
Solve the logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer.
ln left parenthesis x minus 4 right parenthesis plus ln left parenthesis x plus 1 right parenthesis equals ln left parenthesis x minus 8 right parenthesis
Question content area bottom
Part 1
Solve the equation to find the solution set. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet .
(Simplify your answer. Use a comma to separate answers as needed.)
B.
There are infinitely many solutions.
C.
There is no solution.
ln left parenthesis x minus 4 right parenthesis plus ln left parenthesis x plus 1 right parenthesis equals ln left parenthesis x minus 8 right parenthesis
Question content area bottom
Part 1
Solve the equation to find the solution set. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet .
(Simplify your answer. Use a comma to separate answers as needed.)
B.
There are infinitely many solutions.
C.
There is no solution.
Answers
GPT-4o mini
To solve the logarithmic equation
\[
\ln(x - 4) + \ln(x + 1) = \ln(x - 8),
\]
we can use the property of logarithms that states \(\ln(a) + \ln(b) = \ln(ab)\). Applying this property, we can rewrite the left-hand side:
\[
\ln((x - 4)(x + 1)) = \ln(x - 8).
\]
Since the logarithmic function is one-to-one, we can equate the arguments:
\[
(x - 4)(x + 1) = x - 8.
\]
Expanding the left-hand side gives:
\[
x^2 + x - 4x - 4 = x - 8,
\]
which simplifies to
\[
x^2 - 3x - 4 = x - 8.
\]
Now, subtract \(x\) and add \(8\) to both sides:
\[
x^2 - 3x - x - 4 + 8 = 0,
\]
which reduces to:
\[
x^2 - 4x + 4 = 0.
\]
Factoring, we get:
\[
(x - 2)^2 = 0.
\]
Thus, we find the solution:
\[
x = 2.
\]
Now, we need to verify if this solution is in the domain of the original logarithmic expressions. Recall the logarithmic functions are defined for positive arguments:
1. \(x - 4 > 0\) → \(x > 4\)
2. \(x + 1 > 0\) → \(x > -1\)
3. \(x - 8 > 0\) → \(x > 8\)
Since \(x = 2\) does not satisfy \(x > 4\) or \(x > 8\), it is not in the domain of the original logarithmic expressions.
Thus, the final answer is:
**A. The solution set is \(\{\}\) (the empty set).**
\[
\ln(x - 4) + \ln(x + 1) = \ln(x - 8),
\]
we can use the property of logarithms that states \(\ln(a) + \ln(b) = \ln(ab)\). Applying this property, we can rewrite the left-hand side:
\[
\ln((x - 4)(x + 1)) = \ln(x - 8).
\]
Since the logarithmic function is one-to-one, we can equate the arguments:
\[
(x - 4)(x + 1) = x - 8.
\]
Expanding the left-hand side gives:
\[
x^2 + x - 4x - 4 = x - 8,
\]
which simplifies to
\[
x^2 - 3x - 4 = x - 8.
\]
Now, subtract \(x\) and add \(8\) to both sides:
\[
x^2 - 3x - x - 4 + 8 = 0,
\]
which reduces to:
\[
x^2 - 4x + 4 = 0.
\]
Factoring, we get:
\[
(x - 2)^2 = 0.
\]
Thus, we find the solution:
\[
x = 2.
\]
Now, we need to verify if this solution is in the domain of the original logarithmic expressions. Recall the logarithmic functions are defined for positive arguments:
1. \(x - 4 > 0\) → \(x > 4\)
2. \(x + 1 > 0\) → \(x > -1\)
3. \(x - 8 > 0\) → \(x > 8\)
Since \(x = 2\) does not satisfy \(x > 4\) or \(x > 8\), it is not in the domain of the original logarithmic expressions.
Thus, the final answer is:
**A. The solution set is \(\{\}\) (the empty set).**