Logarithm Equation Aa4D7C

1. **State the problem:** Simplify and verify the equation $$2 \log_2 \sqrt{x} + 2 \log_2 y = \log_2 xy$$ and show that it is equivalent to $$\log_2 x + \log_2 y^2 = \log_2 xy$$. 2. **Recall logarithm properties:** - Power rule: $$a \log_b c = \log_b c^a$$ - Product rule: $$\log_b m + \log_b n = \log_b (mn)$$ - Square root as power: $$\sqrt{x} = x^{\frac{1}{2}}$$ 3. **Rewrite the left side of the original equation:** $$2 \log_2 \sqrt{x} + 2 \log_2 y = 2 \log_2 x^{\frac{1}{2}} + 2 \log_2 y$$ 4. **Apply the power rule:** $$2 \log_2 x^{\frac{1}{2}} = \log_2 (x^{\frac{1}{2}})^2 = \log_2 x$$ $$2 \log_2 y = \log_2 y^2$$ 5. **Substitute back:** $$\log_2 x + \log_2 y^2$$ 6. **Use the product rule:** $$\log_2 x + \log_2 y^2 = \log_2 (x \cdot y^2)$$ 7. **Compare with the right side:** The right side is $$\log_2 (xy)$$. 8. **Check equality:** $$\log_2 (x y^2) \neq \log_2 (x y)$$ unless $$y^2 = y$$, which is true only if $$y=1$$ or $$y=0$$. **Therefore, the original equation is not generally true unless $$y=1$$ or $$y=0$$.** **Final answer:** The given equation simplifies to $$\log_2 x + \log_2 y^2 = \log_2 (x y^2)$$ which is not equal to $$\log_2 (x y)$$ in general.