AMC12 2018 A

AMC12 2018 A · Q14

AMC12 2018 A · Q14. It mainly tests Logarithms (rare), Manipulating equations.

The solution to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\frac{1}{3}$ or $\frac{1}{2}$, can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p + q$?

方程$\log_{3x} 4 = \log_{2x} 8$的解（其中$x$是除$\frac{1}{3}$或$\frac{1}{2}$外的正实数）可以写成$\frac{p}{q}$，其中$p$和$q$是互质的正整数。$p + q$等于多少？

(A) 5 5

(B) 13 13

(D) 31 31

(E) 35 35

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): By the change-of-base formula, the given equation is equivalent to $$\frac{\log 4}{\log 3x}=\frac{\log 8}{\log 2x}$$ $$\frac{2\log 2}{\log 3+\log x}=\frac{3\log 2}{\log 2+\log x}$$ $$2\log 2+2\log x=3\log 3+3\log x$$ $$\log x=2\log 2-3\log 3$$ $$\log x=\log \frac{4}{27}.$$ Therefore $x=\frac{4}{27}$, and the requested sum is $4+27=31$.

答案（D）：由换底公式，所给方程等价于 $$\frac{\log 4}{\log 3x}=\frac{\log 8}{\log 2x}$$ $$\frac{2\log 2}{\log 3+\log x}=\frac{3\log 2}{\log 2+\log x}$$ $$2\log 2+2\log x=3\log 3+3\log x$$ $$\log x=2\log 2-3\log 3$$ $$\log x=\log \frac{4}{27}.$$ 因此 $x=\frac{4}{27}$，所求的和为 $4+27=31$。

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