Difference between revisions of "2003 AMC 12B Problems/Problem 17"
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It is not difficult to find <math>x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}</math>. | It is not difficult to find <math>x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}</math>. | ||
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+ | == Solution 2 == | ||
+ | <math>log(xy)+log(y^2)=1 \\ log(xy)+log(x)=1</math> subtracting, <math> \\ log(y^2)-log(x)=0 \\ log \left(\frac{y^2}{x}\right)=0 \\ y^2=x \\ \text{substitute and solve:} 5log(y)=1 \text{ and we need } 3log(y) \text{ which is } \frac{3}{5}</math> | ||
== See also == | == See also == |
Revision as of 21:18, 22 November 2019
Contents
Problem
If and , what is ?
Solution
Since Summing gives
Hence .
It is not difficult to find .
Solution 2
subtracting,
See also
2003 AMC 12B (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 12 Problems and Solutions |
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