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) $\frac{3}{2}$ (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
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1991 AHSME Problems/Problem 20

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