To solve the equation [tex]\(9^{3x+2} = \frac{1}{3^{2x-5}}\)[/tex], we need to manipulate and simplify the exponents. Here's a detailed, step-by-step solution:
1. Rewrite the Equation Using Equivalent Bases:
Notice that [tex]\(9\)[/tex] can be expressed as [tex]\(3^2\)[/tex]. Therefore, we can rewrite the equation as:
[tex]\[
(3^2)^{3x+2} = \frac{1}{3^{2x-5}}
\][/tex]
2. Simplify the Left Side:
Using the power of a power rule [tex]\((a^m)^n = a^{m \cdot n}\)[/tex]:
[tex]\[
3^{2(3x+2)} = \frac{1}{3^{2x-5}}
\][/tex]
Simplifying the exponent on the left-hand side:
[tex]\[
3^{6x + 4} = \frac{1}{3^{2x-5}}
\][/tex]
3. Rewrite the Right Side to Have the Same Base:
The right-hand side can be rewritten using the property of negative exponents [tex]\(\frac{1}{a^m} = a^{-m}\)[/tex]:
[tex]\[
3^{6x + 4} = 3^{-(2x-5)}
\][/tex]
Simplify the exponent on the right-hand side:
[tex]\[
3^{6x + 4} = 3^{-2x + 5}
\][/tex]
4. Equate the Exponents:
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[
6x + 4 = -2x + 5
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
Combine like terms:
[tex]\[
6x + 2x = 5 - 4
\][/tex]
[tex]\[
8x = 1
\][/tex]
Divide both sides by 8:
[tex]\[
x = \frac{1}{8}
\][/tex]
So, one solution is [tex]\(x = \frac{1}{8}\)[/tex].
6. Exploring Complex Exponents:
The right-hand side [tex]\(\frac{1}{3^{2x-5}}\)[/tex] involves negative exponents, indicating possible complex solutions. By carefully solving the complex form of the equation, the solutions are:
[tex]\[
\left[ \frac{1}{8}, \frac{\log(3) - 6i\pi}{8\log(3)}, \frac{\log(3) + 6i\pi}{8\log(3)}, \frac{1}{8} - \frac{i\pi}{2\log(3)}, \frac{1}{8} - \frac{i\pi}{4\log(3)}, \frac{1}{8} + \frac{i\pi}{4\log(3)}, \frac{1}{8} + \frac{i\pi}{2\log(3)}, \frac{1}{8} + \frac{i\pi}{\log(3)} \][/tex]
These are the complete set of solutions, including the complex ones.
Thus, the equation [tex]\(9^{3x+2} = \frac{1}{3^{2x-5}}\)[/tex] has the following solutions:
[tex]\[ x = \frac{1}{8}, \frac{\log(3) - 6i\pi}{8\log(3)}, \frac{\log(3) + 6i\pi}{8\log(3)}, \frac{1}{8} - \frac{i\pi}{2\log(3)}, \frac{1}{8} - \frac{i\pi}{4\log(3)}, \frac{1}{8} + \frac{i\pi}{4\log(3)}, \frac{1}{8} + \frac{i\pi}{2\log(3)}, \frac{1}{8} + \frac{i\pi}{\log(3)} \][/tex]
