Difference between revisions of "2005 AMC 10B Problems/Problem 17"
Dairyqueenxd (talk | contribs) (→Solution) |
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2^3&=2^{2\cdot a\cdot b\cdot c\cdot d}\ | 2^3&=2^{2\cdot a\cdot b\cdot c\cdot d}\ | ||
3&=2\cdot a\cdot b\cdot c\cdot d\ | 3&=2\cdot a\cdot b\cdot c\cdot d\ | ||
− | a\cdot b\cdot c\cdot d&=\boxed{\textbf{(B) } | + | a\cdot b\cdot c\cdot d&=\boxed{\textbf{(B) }\dfrac{3}{2}}\ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 11:56, 16 December 2021
Contents
[hide]Problem
Suppose that , , , and . What is ?
Solution
Solution using logarithms
We can write as , as , as , and as . We know that can be rewritten as , so
Solution using logarithm chain rule
As in solution 2, we can write as , as , as , and as . is equivalent to . Note that by the logarithm chain rule, this is equivalent to , which evaluates to , so is the answer. ~solver1104
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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