Difference between revisions of "1991 AHSME Problems/Problem 20"
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| + | == Problem == | ||
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The sum of all real <math>x</math> such that <math>(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3</math> is | The sum of all real <math>x</math> such that <math>(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3</math> is | ||
(A) 3/2 (B) 2 (C) 5/2 (D) 3 (E) 7/2 | (A) 3/2 (B) 2 (C) 5/2 (D) 3 (E) 7/2 | ||
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| + | == Solution == | ||
| + | <math>\fbox{}</math> | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1991|num-b=19|num-a=21}} | ||
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| + | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:00, 28 September 2014
Problem
The sum of all real 
such that 
is
(A) 3/2 (B) 2 (C) 5/2 (D) 3 (E) 7/2
Solution
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
