Difference between revisions of "1970 AHSME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | Since we want the <math>y</math> expression in terms of <math>x</math>, let's convert the <math>y</math> expression. We can convert it to <math>1+ \frac{1}{2^p} \Rightarrow \frac{2^p+1}{2^p} \Rightarrow \frac{x}{x-1} \Rightarrow</math> <math>\fbox{C}</math> |
== See also == | == See also == | ||
Latest revision as of 23:04, 15 January 2019
Problem
If 
and 
, then 
in terms of 
is
Solution
Since we want the expression in terms of , let's convert the expression. We can convert it to 
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
