Cho hai số thực dương $a, b$. Biết $\log _{2} \frac{\sqrt{4 a^{3} b}}{a b}=1+x \log _{2} a+y \log _{4} b~(x, y \in \mathbb{Q})$. Tính $x+y$

Q: Cho hai số thực dương $a, b$. Biết $\log _{2} \frac{\sqrt{4 a^{3} b}}{a b}=1+x \log _{2} a+y \log _{4} b~(x, y \in \mathbb{Q})$. Tính $x+y$

Ta có: $\log _{2} \frac{\sqrt{4 a^{3} b}}{a b}=\log _{2}\left(2. \frac{a^{\frac{3}{2}}}{a}.\frac{b^{\frac{1}{2}}}{b}\right)$$=\log _{2}\left(2 . a^{\frac{1}{2}} . b^{-\frac{1}{2}}\right)=\log _{2} 2+\log _{2} a^{\frac{1}{2}}+\log _{2} b^{-\frac{1}{2}}=1+\frac{1}{2} \log _{2} a+\left(-\frac{1}{2}\right) \log _{2} b$Vậy $x=\frac{1}{2} ; y=-\frac{1}{2}$ suy ra $x+y=\frac{1}{2}+\left(-\frac{1}{2}\right)=0$

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Accepted Answer

Ta có: $\log _{2} \frac{\sqrt{4 a^{3} b}}{a b}=\log _{2}\left(2. \frac{a^{\frac{3}{2}}}{a}.\frac{b^{\frac{1}{2}}}{b}\right)$

$=\log _{2}\left(2 . a^{\frac{1}{2}} . b^{-\frac{1}{2}}\right)=\log _{2} 2+\log _{2} a^{\frac{1}{2}}+\log _{2} b^{-\frac{1}{2}}=1+\frac{1}{2} \log _{2} a+\left(-\frac{1}{2}\right) \log _{2} b$

Vậy $x=\frac{1}{2} ; y=-\frac{1}{2}$ suy ra $x+y=\frac{1}{2}+\left(-\frac{1}{2}\right)=0$

Đề thi cuối kì 2 (Cấu trúc mới) - Đề số 19 - MĐ 10966