(b) Given that \( \log _{27} x^{6}=\log _{81} y^{2} \), express \( y \) in terms of \( x \).

To solve the equation \( \log_{27} x^{6} = \log_{81} y^{2} \), we can start by rewriting the logarithms in terms of base 3, since both 27 and 81 are powers of 3. First, we note that: \[ 27 = 3^3 \quad \text{and} \quad 81 = 3^4. \] Using the change of base formula for logarithms, we can express the logarithms as follows: \[ \log_{27} x^{6} = \frac{\log_{3} x^{6}}{\log_{3} 27} = \frac{\log_{3} x^{6}}{3} = \frac{6 \log_{3} x}{3} = 2 \log_{3} x. \] Similarly, for \( \log_{81} y^{2} \): \[ \log_{81} y^{2} = \frac{\log_{3} y^{2}}{\log_{3} 81} = \frac{\log_{3} y^{2}}{4} = \frac{2 \log_{3} y}{4} = \frac{1}{2} \log_{3} y. \] Now we can set the two expressions equal to each other: \[ 2 \log_{3} x = \frac{1}{2} \log_{3} y. \] To eliminate the fraction, we can multiply both sides by 2: \[ 4 \log_{3} x = \log_{3} y. \] Next, we can exponentiate both sides to remove the logarithm: \[ y = 3^{4 \log_{3} x}. \] Using the property of exponents, \( a^{\log_{a} b} = b \), we can simplify \( 3^{4 \log_{3} x} \): \[ y = x^4. \] Thus, we have expressed \( y \) in terms of \( x \): \[ \boxed{y = x^4}. \]