To solve the equation [tex]\(\left(\frac{1}{27}\right)^{2-x}=9^{3 x}\)[/tex], let’s follow these steps:
1. Rewrite the bases in terms of powers of 3:
[tex]\[
\frac{1}{27} = \left(\frac{1}{3^3}\right) = 3^{-3}
\][/tex]
[tex]\[
9 = 3^2
\][/tex]
2. Substitute these rewritten bases into the equation:
[tex]\[
\left(3^{-3}\right)^{2-x} = \left(3^2\right)^{3x}
\][/tex]
3. Simplify the exponents:
When raising a power to another power, you multiply the exponents:
[tex]\[
3^{-3(2-x)} = 3^{2 \cdot 3x}
\][/tex]
[tex]\[
3^{-6 + 3x} = 3^{6x}
\][/tex]
4. Since the bases are the same, set the exponents equal to each other:
[tex]\[
-6 + 3x = 6x
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[
-6 + 3x = 6x
\][/tex]
[tex]\[
-6 = 6x - 3x
\][/tex]
[tex]\[
-6 = 3x
\][/tex]
[tex]\[
x = \frac{-6}{3}
\][/tex]
[tex]\[
x = -2
\][/tex]
Therefore, the correct answer is:
A. -2
