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.
log Subscript 4 Baseline left parenthesis x plus 6 right parenthesis plus log Subscript 4 Baseline left parenthesis x plus 3 right parenthesis equals 1
Question content area bottom
Part 1
Solve the equation. 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.
log Subscript 4 Baseline left parenthesis x plus 6 right parenthesis plus log Subscript 4 Baseline left parenthesis x plus 3 right parenthesis equals 1
Question content area bottom
Part 1
Solve the equation. 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
\[
\log_4 (x + 6) + \log_4 (x + 3) = 1,
\]
we first use the property of logarithms that states \(\log_a b + \log_a c = \log_a (bc)\). This allows us to combine the logs:
\[
\log_4 \left((x + 6)(x + 3)\right) = 1.
\]
Next, we can rewrite the equation in exponential form. Since \( \log_4 A = B\) implies \(A = 4^B\), we get:
\[
(x + 6)(x + 3) = 4^1,
\]
which simplifies to:
\[
(x + 6)(x + 3) = 4.
\]
Now, we can expand the left-hand side:
\[
x^2 + 3x + 6x + 18 = 4,
\]
which simplifies to:
\[
x^2 + 9x + 18 = 4.
\]
Next, we subtract 4 from both sides to set the equation to zero:
\[
x^2 + 9x + 14 = 0.
\]
Now we can factor the quadratic equation:
\[
(x + 7)(x + 2) = 0.
\]
Setting each factor to zero gives us the possible solutions:
\[
x + 7 = 0 \quad \Rightarrow \quad x = -7,
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2.
\]
Next, we need to check the domain of the original logarithmic expressions. The arguments of the logarithms must be positive:
1. For \(\log_4 (x + 6)\): \( x + 6 > 0 \) implies \( x > -6 \).
2. For \(\log_4 (x + 3)\): \( x + 3 > 0 \) implies \( x > -3 \).
From the domain restrictions, the solution \(x = -7\) is not valid because it does not satisfy \(x > -6\) or \(x > -3\).
The solution \(x = -2\) does satisfy the domain conditions:
\(-2 > -6\) and \(-2 > -3\).
Thus, the only valid solution is:
\[
x = -2.
\]
Finally, since we do have a solution in the valid domain, we conclude that the correct choice is:
**A. The solution set is \(\{-2\}\).**
\[
\log_4 (x + 6) + \log_4 (x + 3) = 1,
\]
we first use the property of logarithms that states \(\log_a b + \log_a c = \log_a (bc)\). This allows us to combine the logs:
\[
\log_4 \left((x + 6)(x + 3)\right) = 1.
\]
Next, we can rewrite the equation in exponential form. Since \( \log_4 A = B\) implies \(A = 4^B\), we get:
\[
(x + 6)(x + 3) = 4^1,
\]
which simplifies to:
\[
(x + 6)(x + 3) = 4.
\]
Now, we can expand the left-hand side:
\[
x^2 + 3x + 6x + 18 = 4,
\]
which simplifies to:
\[
x^2 + 9x + 18 = 4.
\]
Next, we subtract 4 from both sides to set the equation to zero:
\[
x^2 + 9x + 14 = 0.
\]
Now we can factor the quadratic equation:
\[
(x + 7)(x + 2) = 0.
\]
Setting each factor to zero gives us the possible solutions:
\[
x + 7 = 0 \quad \Rightarrow \quad x = -7,
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2.
\]
Next, we need to check the domain of the original logarithmic expressions. The arguments of the logarithms must be positive:
1. For \(\log_4 (x + 6)\): \( x + 6 > 0 \) implies \( x > -6 \).
2. For \(\log_4 (x + 3)\): \( x + 3 > 0 \) implies \( x > -3 \).
From the domain restrictions, the solution \(x = -7\) is not valid because it does not satisfy \(x > -6\) or \(x > -3\).
The solution \(x = -2\) does satisfy the domain conditions:
\(-2 > -6\) and \(-2 > -3\).
Thus, the only valid solution is:
\[
x = -2.
\]
Finally, since we do have a solution in the valid domain, we conclude that the correct choice is:
**A. The solution set is \(\{-2\}\).**