For all positive numbers x distinct from 1, 1log_3x+1log_4x+1log_5x equals

(D) (12)/((log_3x)+(log_4x)+(log_5x))

(E) (log_2x)/((log_3x)(log_5x)) +(log_3x)/((log_2x)(log_5x)) +(log_5x)/((log_2x)(log_3x))

1978 AMC 12 Problem 21

Attempt Problem 21 of the 1978 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1978 AMC 12 solutions, or check the answer key.

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21.

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)}

Use the reciprocal identity

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

Combine the resulting sum with the product rule for logarithms

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.