1. **State the problem:** Solve the equation $$\log_2(y^4) + \log_2(y) - \log_2(4y) = 2$$ for $y$. 2. **Recall logarithm properties:** - $\log_b(a) + \log_b(c) = \log_b(ac)$ - $\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)$ - $\log_b(a^n) = n\log_b(a)$ 3. **Apply the properties:** Combine the first two terms: $$\log_2(y^4) + \log_2(y) = \log_2(y^4 \cdot y) = \log_2(y^5)$$ So the equation becomes: $$\log_2(y^5) - \log_2(4y) = 2$$ 4. **Combine the logarithms:** $$\log_2\left(\frac{y^5}{4y}\right) = 2$$ Simplify inside the log: $$\frac{y^5}{4y} = \frac{y^{5-1}}{4} = \frac{y^4}{4}$$ So: $$\log_2\left(\frac{y^4}{4}\right) = 2$$ 5. **Rewrite the logarithmic equation in exponential form:** $$\frac{y^4}{4} = 2^2$$ $$\frac{y^4}{4} = 4$$ 6. **Solve for $y^4$:** Multiply both sides by 4: $$\cancel{4} \cdot \frac{y^4}{\cancel{4}} = 4 \cdot 4$$ $$y^4 = 16$$ 7. **Solve for $y$:** Take the fourth root of both sides: $$y = \pm \sqrt[4]{16}$$ Since $16 = 2^4$, $$y = \pm 2$$ 8. **Check domain restrictions:** The original logarithms require arguments to be positive: - $y^4 > 0$ always true for $y \neq 0$ - $y > 0$ so $y$ must be positive - $4y > 0$ so $y > 0$ Therefore, $y$ must be positive. 9. **Final answer:** $$\boxed{y = 2}$$