solve: $243^{-y}=\\left(\\frac{1}{243}\\right)^{3y}\\cdot9^{-2y}$\n$y = - 1$\n$y = 0$\n$y = 1$\nno solution

Explanation:

Step1: Rewrite bases as powers of 3

Since (243 = 3^5) and (9=3^2), the equation (243^{-y}=\left(\frac{1}{243}\right)^{3y}\cdot9^{-2y}) can be rewritten as ((3^5)^{-y}=(3^{- 5})^{3y}\cdot(3^2)^{-2y}).

Step2: Apply power - of - a - power rule

Using ((a^m)^n=a^{mn}), we get (3^{-5y}=3^{-15y}\cdot3^{-4y}).

Step3: Apply product rule of exponents

Since (a^m\cdot a^n=a^{m + n}), the right - hand side is (3^{-15y-4y}=3^{-19y}). So the equation becomes (3^{-5y}=3^{-19y}).

Step4: Set exponents equal

If (a^m=a^n), then (m = n). So, (-5y=-19y).

Step5: Solve for y

Add (19y) to both sides: (-5y + 19y=-19y+19y), which gives (14y = 0). Then (y = 0).

Answer:

(y = 0)