To determine which equation is equivalent to [tex]\(\left(\frac{1}{3}\right)^x = 27^{x+2}\)[/tex], let's follow a step-by-step solution:
1. Rewrite the bases using powers of 3:
- The term [tex]\(\frac{1}{3}\)[/tex] can be written as [tex]\(3^{-1}\)[/tex].
- The number 27 can be expressed as [tex]\(3^3\)[/tex].
Therefore, we can rewrite the equation [tex]\(\left(\frac{1}{3}\right)^x\)[/tex] and [tex]\(27^{x+2}\)[/tex] in terms of base 3:
[tex]\[
\left(3^{-1}\right)^x = \left(3^3\right)^{x+2}
\][/tex]
2. Apply the power rule [tex]\( \left(a^m\right)^n = a^{m \cdot n} \)[/tex]:
- For the left side: [tex]\(\left(3^{-1}\right)^x = 3^{-1 \cdot x} = 3^{-x}\)[/tex]
- For the right side: [tex]\(\left(3^3\right)^{x+2} = 3^{3 \cdot (x+2)} = 3^{3x + 6}\)[/tex]
So the equation becomes:
[tex]\[
3^{-x} = 3^{3x + 6}
\][/tex]
3. Equate the exponents since the bases are the same:
- If [tex]\(3^{-x} = 3^{3x + 6}\)[/tex], then:
[tex]\[
-x = 3x + 6
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
-x - 3x = 6
\][/tex]
[tex]\[
-4x = 6
\][/tex]
[tex]\[
x = -\frac{3}{2}
\][/tex]
Now, based on this work, we have identified the equivalent equation in terms of the same base (3). The equivalent equation is:
[tex]\[
3^{-x} = 3^{3x + 6}
\][/tex]
Thus, the correct answer is:
[tex]\[
3^{-x} = 3^{3x + 6}
\][/tex]
So, the equivalent equation is:
[tex]\[
3^{-x} = 3^{3x + 6}
\][/tex]
