1978 AHSME Problems/Problem 21

Problem 21

For all positive numbers $x$ distinct from $1$ ,

$\frac{1}{\log_3(x)}+\frac{1}{\log_4(x)}+\frac{1}{\log_5(x)}$

equals

$\text{(A) }\frac{1}{\log_{60}(x)}\qquad\\ \text{(B) }\frac{1}{\log_{x}(60)}\qquad\\ \text{(C) }\frac{1}{(\log_{3}(x))(\log_{4}(x))(\log_{5}(x))}\qquad\\ \text{(D) }\frac{12}{\log_{3}(x)+\log_{4}(x)+\log_{5}(x)}\qquad\\ \text{(E) }\frac{\log_{2}(x)}{\log_{3}(x)\log_{5}(x)}+\frac{\log_{3}(x)}{\log_{2}(x)\log_{5}(x)}+\frac{\log_{5}(x)}{\log_{2}(x)\log_{3}(x)}$

Solution

$\frac{1}{\log_3(x)}+\frac{1}{\log_4(x)}+\frac{1}{\log_5(x)}\implies \frac{1}{\frac{\log(x)}{\log(3)}}+\frac{1}{\frac{\log(x)}{\log(4)}}+\frac{1}{\frac{\log(x)}{\log(5)}}\implies \frac{\log(3)+\log(4)+\log(5)}{\log(x)}=\frac{\log(60)}{\log(x)}\implies \log_x(60)\implies \frac{1}{\log_{60}(x)}$ Thus, the answer is $(\textbf{A})$

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1978 AHSME Problems/Problem 21

Problem 21

Solution