AMC12 2003 B

AMC12 2003 B · Q17

AMC12 2003 B · Q17. It mainly tests Logarithms (rare).

If $\log (xy^3) = 1$ and $\log (x^2y) = 1$, what is $\log (xy)$?

若 $\log (xy^3) = 1$ 且 $\log (x^2y) = 1$，求 $\log (xy)$？

(A) $-\frac{1}{2}$ $-\frac{1}{2}$

(B) 0 0

(D) $\frac{3}{5}$ $\frac{3}{5}$

(E) 1 1

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since \begin{align*} &\log(xy) +2\log y = 1 \\ \log(xy) + \log x = 1 \quad \Longrightarrow \quad &2\log(xy) + 2\log x = 2 \end{align*} Summing gives \[3\log(xy) + 2\log y + 2\log x = 3 \Longrightarrow 5\log(xy) = 3\] Hence $\log (xy) = \frac 35 \Rightarrow \mathrm{(D)}$. It is not difficult to find $x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}$.

因为 \begin{align*} &\log(xy) +2\log y = 1 \\ \log(xy) + \log x = 1 \quad \Longrightarrow \quad &2\log(xy) + 2\log x = 2 \end{align*} 相加得到 \[3\log(xy) + 2\log y + 2\log x = 3 \Longrightarrow 5\log(xy) = 3\] 因此 $\log (xy) = \frac 35 \Rightarrow \mathrm{(D)}$。不难求得 $x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}$。

Topics

AL.015 Logarithms (rare)