Difference between revisions of "1991 AHSME Problems/Problem 20"

Revision as of 16:30, 27 October 2018

Problem

The sum of all real $x$ such that $(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3$ is

(A) $\frac{3}{2}$ (B) $2$ (C) $\frac{5}{2}$ (D) $3$ (E) $\frac{7}{2}$

Solution

$\fbox{E}$ Let $a = 2^x-4, b = 4^x-2$ , so the equation becomes $a^3+b^3=(a+b)^3 = a^3+3a^2b+3ab^2+b^3 \implies 3a^2b+3ab^2=0 \implies ab(a+b)=0$ . Hence $a=0 \implies 2^x=4 \implies x=2$ , or $b=0 \implies 4^x=2 \implies x=\frac{1}{2}$ , or $a+b=0 \implies 4^x+2^x-6=0 \implies (2^x)^2+2^x-6=0 \implies (2^x+3)(2^x-2)=0$ , but $2^x >0$ for all $x$ , so we cannot have $2^x = -3$ ; however, $2^x=2$ works, giving $x=1.$ Thus we have $3$ solutions: $2, 1/2, 1$ , whose sum is $\frac{7}{2}.$

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.

@@ Line 3: / Line 3: @@
 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) <math>\frac{3}{2}</math>  (B) <math>2</math>  (C) <math>\frac{5}{2}</math>  (D) <math>3</math>  (E) <math>\frac{7}{2}</math>
 == Solution ==

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Difference between revisions of "1991 AHSME Problems/Problem 20"

Revision as of 16:30, 27 October 2018

Problem

Solution

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