Cho $\Large \dfrac{\mathrm{log}a}{p}=\dfrac{\mathrm{log}b}{q}=\dfrac{\

Cho $\Large \dfrac{\mathrm{log}a}{p}=\dfrac{\mathrm{log}b}{q}=\dfrac{\mathrm{log}c}{r}=\mathrm{log}x\neq 0$; $\Large \dfrac{b^2}{ac}=x^y$. Tính $\Large y$ theo $\Large p, q, r$.

• A.

  $\Large y=q^2-pr$.

• B.

  $\Large y=\dfrac{p+r}{2q}$.

• C.

  $\Large y=2q-p-r$.

• D.

  $\Large y=2q-pr$.

Chọn C
+) Ta có $\Large \dfrac{b^2}{ac}=x^y$ $\Large \Rightarrow y\mathrm{log}x^y=\mathrm{log}\dfrac{b^2}{ac}=\mathrm{log}b^2-\mathrm{log}ac$ $\Large =2\mathrm{log}b-\mathrm{log}a-\mathrm{log}c$

$\Large \Rightarrow y=\dfrac{2\mathrm{log}b}{\mathrm{log}x}-\dfrac{\mathrm{log}a}{\mathrm{log}x}-\dfrac{\mathrm{log}c}{\mathrm{log}x}$ (vì $\Large \mathrm{log}x\neq 0$).

+) Từ đó suy ra $\Large y=2q-p-r$.