which equation is equivalent to $\\left(\\frac{1}{3}\\right)^{x}=27^{x + 2}$?\n$3^{x}=3^{-3x +…

Explanation:

Step1: Rewrite $\frac{1}{3}$ as a power of 3

Since $\frac{1}{3}=3^{- 1}$, then $(\frac{1}{3})^x=(3^{-1})^x$. By the power - of - a - power rule $(a^m)^n=a^{mn}$, we have $(3^{-1})^x = 3^{-x}$.

Step2: Rewrite 27 as a power of 3

Since $27 = 3^3$, then $27^{x + 2}=(3^3)^{x+2}$. Using the power - of - a - power rule $(a^m)^n=a^{mn}$, we get $(3^3)^{x + 2}=3^{3(x + 2)}$.

Step3: Expand $3(x + 2)$

By the distributive property $a(b + c)=ab+ac$, $3(x + 2)=3x+6$. So $27^{x + 2}=3^{3x + 6}$.

Step4: Write the equivalent equation

The original equation $(\frac{1}{3})^x=27^{x + 2}$ is equivalent to $3^{-x}=3^{3x + 6}$.

Answer:

$3^{-x}=3^{3x + 6}$