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 $\frac{7}{2}.$

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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