If (0.25)\(^y\) = 32, find the value of y.

We can solve the given equation by taking the logarithm of both sides. Any base of the logarithm can be used, but we will use the common logarithm (base 10). log\((0.25)^y\) = log 32 Using the logarithmic identity log\((a^b)\) = b log\((a)\), we get: y log\((0.25)\) = log 32 Now, we can evaluate log\((0.25)\) using the logarithmic identity log\((a^n)\) = n log\((a)\), as follows: log\((0.25)\) = log\((\frac{1}{4})\) = log\((4^{-1})\) = -log\((4)\) We know that log\((10^x)\) = x, so log\((4)\) = log\((10^{0.602})\) \(\approx\) 0.602 Therefore, y log\((0.25)\) = log 32 y (-log\((4)\)) = log 32 y (-0.602) = log 32 y = \(\frac{\text{log } 32}{-0.602}\) Using a calculator, we get: y \(\approx\) -2.5 Therefore, the value of y is approximately -2.5. Hence, the answer is: y = -\(\frac{5}{2}\).