Question
Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expression. Give the exact answer.
log Subscript 2 Baseline left parenthesis x plus 3 right parenthesis equals negative 4
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
.
(Type an exact answer in simplified form. Use integers or fractions for any numbers in the expression.)
B.
There are infinitely many solutions.
C.
There is no solution.
log Subscript 2 Baseline left parenthesis x plus 3 right parenthesis equals negative 4
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
.
(Type an exact answer in simplified form. Use integers or fractions for any numbers in the expression.)
B.
There are infinitely many solutions.
C.
There is no solution.
Answers
GPT-4o mini
To solve the equation \( \log_2(x + 3) = -4 \), we can express this logarithmic equation in its exponential form. The equation \( \log_b(a) = c \) translates to \( a = b^c \).
1. Exponential form:
\[
x + 3 = 2^{-4}
\]
2. Calculate \( 2^{-4} \):
\[
2^{-4} = \frac{1}{16}
\]
3. Substitute back into the equation:
\[
x + 3 = \frac{1}{16}
\]
4. Solve for \( x \):
\[
x = \frac{1}{16} - 3
\]
5. Convert 3 into a fraction:
\[
3 = \frac{48}{16} \quad \text{so} \quad x = \frac{1}{16} - \frac{48}{16} = \frac{1 - 48}{16} = \frac{-47}{16}
\]
Next, we need to check if this value \( x \) is in the domain of the original logarithmic expression.
The expression \( \log_2(x + 3) \) is defined only when \( x + 3 > 0 \).
6. Check the condition:
\[
x + 3 > 0 \implies \frac{-47}{16} + 3 > 0
\]
Convert 3 to a fraction:
\[
3 = \frac{48}{16} \implies \frac{-47}{16} + \frac{48}{16} > 0 \implies \frac{1}{16} > 0
\]
This condition is true.
Since \( \frac{-47}{16} + 3 = \frac{1}{16} \) is positive, this means that \( x = \frac{-47}{16} \) is in the domain of the original logarithmic expression and is a valid solution.
Thus, the solution is:
\[
x = \frac{-47}{16}
\]
Now, referring to the options provided in the question:
A. The solution set is \( \left\{ \frac{-47}{16} \right\} \)
B. There are infinitely many solutions.
C. There is no solution.
The correct choice is **A** with the solution \( x = \frac{-47}{16} \).
1. Exponential form:
\[
x + 3 = 2^{-4}
\]
2. Calculate \( 2^{-4} \):
\[
2^{-4} = \frac{1}{16}
\]
3. Substitute back into the equation:
\[
x + 3 = \frac{1}{16}
\]
4. Solve for \( x \):
\[
x = \frac{1}{16} - 3
\]
5. Convert 3 into a fraction:
\[
3 = \frac{48}{16} \quad \text{so} \quad x = \frac{1}{16} - \frac{48}{16} = \frac{1 - 48}{16} = \frac{-47}{16}
\]
Next, we need to check if this value \( x \) is in the domain of the original logarithmic expression.
The expression \( \log_2(x + 3) \) is defined only when \( x + 3 > 0 \).
6. Check the condition:
\[
x + 3 > 0 \implies \frac{-47}{16} + 3 > 0
\]
Convert 3 to a fraction:
\[
3 = \frac{48}{16} \implies \frac{-47}{16} + \frac{48}{16} > 0 \implies \frac{1}{16} > 0
\]
This condition is true.
Since \( \frac{-47}{16} + 3 = \frac{1}{16} \) is positive, this means that \( x = \frac{-47}{16} \) is in the domain of the original logarithmic expression and is a valid solution.
Thus, the solution is:
\[
x = \frac{-47}{16}
\]
Now, referring to the options provided in the question:
A. The solution set is \( \left\{ \frac{-47}{16} \right\} \)
B. There are infinitely many solutions.
C. There is no solution.
The correct choice is **A** with the solution \( x = \frac{-47}{16} \).