Difference between revisions of "1978 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
| − | <cmath>\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}{\ | + | <cmath>\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)}</cmath> |
Thus, the answer is <math>(\textbf{A})</math> | Thus, the answer is <math>(\textbf{A})</math> | ||
Latest revision as of 09:32, 19 May 2016
Problem 21
For all positive numbers 
distinct from 
,
![\[\frac{1}{\log_3(x)}+\frac{1}{\log_4(x)}+\frac{1}{\log_5(x)}\]](http://latex.artofproblemsolving.com/b/4/d/b4dd56ee1fad69f839cd4244cdedf9dc5a7e85ab.png)
equals
Solution
![\[\frac{1}{\log_3(x)}+\frac{1}{\log_4(x)}+\frac{1}{\log_5(x)}\implies \frac{1}{\frac{\log(x)}{\log(3)}}+\frac{1}{\frac{\log(x)}{\log(4)}}+\frac{1}{\frac{\log(x)}{\log(5)}}\implies \frac{\log(3)+\log(4)+\log(5)}{\log(x)}=\frac{\log(60)}{\log(x)}\implies \log_x(60)\implies \frac{1}{\log_{60}(x)}\]](http://latex.artofproblemsolving.com/f/2/b/f2b3bf3d28d3a3ca49d73a74368095731f0a6aa7.png)
Thus, the answer is 
