1. **Stating the problem**: Given $\log_4 x = y$, where $x > 0$, find: a) $\log_2 x$ b) $\log_8 x$ c) $\log_{16} x$ 2. **Express $x$ in terms of $y$:** $$\log_4 x = y \implies x = 4^y$$ 3. **Rewrite $4$ in base $2$:** $$4 = 2^2$$ So, $$x = (2^2)^y = 2^{2y}$$ 4. **Compute each requested logarithm:** a) $\log_2 x = \log_2 (2^{2y}) = 2y$ b) Rewrite $8$ as $2^3$, then use change of base: $$\log_8 x = \log_8 (2^{2y}) = \frac{\log_2 (2^{2y})}{\log_2 8} = \frac{2y}{3}$$ c) Rewrite $16$ as $2^4$, then similarly: $$\log_{16} x = \log_{16} (2^{2y}) = \frac{\log_2 (2^{2y})}{\log_2 16} = \frac{2y}{4} = \frac{y}{2}$$ **Final answers:** a) $\log_2 x = 2y$ b) $\log_8 x = \frac{2y}{3}$ c) $\log_{16} x = \frac{y}{2}$