Sure, let's solve the equation step-by-step:
Given equation:
[tex]\[ 3^{2x - 2} - 3^{2x - 1} = -6 \][/tex]
1. Rewrite the equation using substitution:
Let's use a substitution to simplify the equation. Set:
[tex]\[ y = 3^{2x - 2} \][/tex]
Next, notice that:
[tex]\[ 3^{2x - 1} = 3 \cdot 3^{2x - 2} = 3y \][/tex]
Substituting these back into the original equation, we get:
[tex]\[ y - 3y = -6 \][/tex]
2. Simplify the equation:
Combine the [tex]\(y\)[/tex] terms on the left side:
[tex]\[ -2y = -6 \][/tex]
3. Solve for [tex]\(y\)[/tex]:
Divide both sides of the equation by [tex]\(-2\)[/tex]:
[tex]\[ y = 3 \][/tex]
4. Substitute back to find [tex]\(x\)[/tex]:
Recall that [tex]\( y = 3^{2x - 2} \)[/tex]:
[tex]\[ 3 = 3^{2x - 2} \][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[ 2x - 2 = 1 \][/tex]
5. Solve for [tex]\(x\)[/tex]:
Add 2 to both sides:
[tex]\[ 2x = 3 \][/tex]
Divide both sides by 2:
[tex]\[ x = \frac{3}{2} \][/tex]
6. Identify all possible solutions:
There are also complex solutions due to the nature of exponential equations. The general solution for [tex]\(3^{2x - 2} = 3\)[/tex] includes an additional complex solution. This solution can be written using logarithms:
[tex]\[ 3^{2x - 2} = 3 \Rightarrow 3^{2x - 2} = 3^1 \Rightarrow 2x - 2 = 1 + 2k\pi i / \log(3) \][/tex] for integers [tex]\(k\)[/tex].
This complex solution simplifies further to:
[tex]\[ x = \frac{\log(27) + 2k\pi i}{2 \log(3)} \][/tex]
For [tex]\(k = 0\)[/tex], you get:
[tex]\[ x = \frac{\log(27) + 0\cdot 2\pi i}{2 \log(3)} = \frac{\log(27)}{2 \log(3)} = \frac{3 \log(3)}{2 \log(3)} = \frac{3}{2} \][/tex]
For [tex]\(k = 1\)[/tex], you get:
[tex]\[ x = \frac{\log(27) + 2\pi i}{2 \log(3)} \][/tex]
Therefore, the complete solutions are:
[tex]\[
x = \frac{3}{2} \quad \text{and} \quad x = \frac{\log(27)/2 + i\pi}{\log(3)}
\][/tex]
These match the solutions provided. Hence, the solutions to the equation [tex]\(3^{2x - 2} - 3^{2x - 1} = -6\)[/tex] are:
[tex]\[
x = \frac{3}{2} \quad \text{or} \quad x = \frac{\log(27)}{2 \log(3)} + \frac{i\pi}{\log(3)}
\][/tex]
