1998 AHSME Problems/Problem 22
Problem
What is the value of the expression
Solution
Solution 1
By the change-of-base formula, Thus (you might recognize this identity directly) Thus the sum is
Solution 2
Since ,
$$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{\log_{k}100!}=\frac{\log_{k}k}{\log_{k}100!}=\log_{100!} k\log_{100!} 1 +\log_{100!} 1 +\log_{100!} 1 +\cdots + \log_{100!} 100=\log_{100!}100!=1 \Rightarrow \mathrm{(C)}$
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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