1991 AHSME Problems/Problem 20

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) 3/2 (B) 2 (C) 5/2 (D) 3 (E) 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.

Page

Toolbox

Search

1991 AHSME Problems/Problem 20

Problem

Solution

See also