1978 AMC 12 第 21 题

Question

对于所有不等于

1

的正数

x

，

\frac1{\log_3x}+\frac1{\log_4x}+\frac1{\log_5x}

等于

For all positive numbers

x

distinct from

1,

\frac1{\log_3x}+\frac1{\log_4x}+\frac1{\log_5x}

equals

\frac1{\log_{60}x}

\frac1{\log_x60}

\frac1{(\log_3x)(\log_4x)(\log_5x)}

\frac{12}{(\log_3x)+(\log_4x)+(\log_5x)}

\frac{\log_2x}{(\log_3x)(\log_5x)}

{}+\frac{\log_3x}{(\log_2x)(\log_5x)}

{}+\frac{\log_5x}{(\log_2x)(\log_3x)}

利用倒数恒等式

\frac{1}{\log_bx}=\log_xb

Use the reciprocal identity

\frac{1}{\log_bx}=\log_xb

利用对数的乘积法则合并所得和式

Combine the resulting sum with the product rule for logarithms

令

S

表示题给和式。换底并合并对数可得

\begin{aligned} S&=\frac1{\log_3x}+\frac1{\log_4x}\\ &\quad+\frac1{\log_5x}\\ &=\log_x(3\cdot4\cdot5)\\ &=\log_x60\\ &=\frac1{\log_{60}x}\text{。} \end{aligned}

因此，正确答案是 A。

Let

S

denote the given sum. Changing bases and combining logarithms gives

\begin{aligned} S&=\frac1{\log_3x}+\frac1{\log_4x}\\ &\quad+\frac1{\log_5x}\\ &=\log_x(3\cdot4\cdot5)\\ &=\log_x60\\ &=\frac1{\log_{60}x}. \end{aligned}

Therefore, the correct answer is A.

其他年份的第 21 题