Solve. \( \log _{5}(y)=-3 \) \( y=\square \)

To solve for \( y \) in the equation \( \log_{5}(y) = -3 \), we need to rewrite it in exponential form. This means we can express it as: \[ y = 5^{-3} \] Calculating \( 5^{-3} \), we find: \[ y = \frac{1}{5^3} = \frac{1}{125} \] So, \( y = \frac{1}{125} \) Now you have your final answer! Keep in mind that logarithms can transform multiplication and division into addition and subtraction, which is super handy in mathematics! Just think of them as the bridge between exponential and arithmetic worlds! In the realm of logarithms, many students mistakenly think they can treat the base or the argument as plain numbers without considering their properties, which can lead to confusion. Always remember that the logarithm "undoes" the exponent, but the base used must be consistent. Get comfortable with this concept, and you’ll be breezing through logarithmic equations in no time!