for what value of y does 125 = (\\frac{1}{25})^{y - 1}?\n-\\frac{5}{2}\n-2\n-1\n-\\frac{1}{2}

Explanation:

Step1: Rewrite bases

Rewrite (125) as (5^{3}) and (\frac{1}{25}) as (5^{- 2}). So the equation (125 = (\frac{1}{25})^{y - 1}) becomes (5^{3}=(5^{-2})^{y - 1}).

Step2: Apply power - of - a - power rule

According to the power - of - a - power rule ((a^{m})^{n}=a^{mn}), so ((5^{-2})^{y - 1}=5^{-2(y - 1)}). The equation is now (5^{3}=5^{-2(y - 1)}).

Step3: Set exponents equal

Since the bases are the same ((a^{m}=a^{n}) implies (m = n) for (a>0,a\neq1)), we have (3=-2(y - 1)).

Step4: Solve for y

Expand the right - hand side: (3=-2y + 2). Subtract 2 from both sides: (3-2=-2y), which gives (1=-2y). Then divide both sides by (-2) to get (y =-\frac{1}{2}).

Answer:

(-\frac{1}{2})