Question
Solve the following exponential equation by expressing each side as a power of the same base and then equating exponents.
3 Superscript 2 x minus 12 Baseline equals 9
Question content area bottom
Part 1
The solution set is StartSet nothing EndSet .
3 Superscript 2 x minus 12 Baseline equals 9
Question content area bottom
Part 1
The solution set is StartSet nothing EndSet .
Answers
GPT-4o mini
To solve the equation \( 3^{2x} - 12 = 9 \), we start by isolating the exponential term.
1. **Rearranging the equation:**
\[
3^{2x} = 9 + 12
\]
\[
3^{2x} = 21
\]
2. **Expressing both sides as powers of 3:**
We know that \( 9 = 3^2 \). However, \( 21 \) cannot be expressed as a power of \( 3 \).
Thus, we have:
\[
3^{2x} = 21
\]
3. **Taking logarithms:**
Since \( 21 \) is not a power of \( 3 \), we take the logarithm of both sides.
\[
2x = \log_3(21)
\]
To find \( x \):
\[
x = \frac{1}{2} \log_3(21)
\]
The solution set is not empty, and the answer does not depend on expressing both sides as the same base, since \( 21 \) cannot be written as a power of \( 3 \).
4. **Final result:**
The solution is \( x = \frac{1}{2} \log_3(21) \).
However, if we strictly want the solution set in relation to your original statement,
the result is that there is still a solution, which isn't an empty set.
The solution set is:
\[
\{ x = \frac{1}{2} \log_3(21) \}
\]
If you intended to express no real number solutions due to the base constraints, the statement would have been inaccurate. There is a valid solution in logarithmic terms.
1. **Rearranging the equation:**
\[
3^{2x} = 9 + 12
\]
\[
3^{2x} = 21
\]
2. **Expressing both sides as powers of 3:**
We know that \( 9 = 3^2 \). However, \( 21 \) cannot be expressed as a power of \( 3 \).
Thus, we have:
\[
3^{2x} = 21
\]
3. **Taking logarithms:**
Since \( 21 \) is not a power of \( 3 \), we take the logarithm of both sides.
\[
2x = \log_3(21)
\]
To find \( x \):
\[
x = \frac{1}{2} \log_3(21)
\]
The solution set is not empty, and the answer does not depend on expressing both sides as the same base, since \( 21 \) cannot be written as a power of \( 3 \).
4. **Final result:**
The solution is \( x = \frac{1}{2} \log_3(21) \).
However, if we strictly want the solution set in relation to your original statement,
the result is that there is still a solution, which isn't an empty set.
The solution set is:
\[
\{ x = \frac{1}{2} \log_3(21) \}
\]
If you intended to express no real number solutions due to the base constraints, the statement would have been inaccurate. There is a valid solution in logarithmic terms.