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

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== Problem ==
 
== 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
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<math>\textbf{(A) } \frac32 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } \frac52 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \frac72</math>
  
 
== Solution ==
 
== Solution ==
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<math>\fbox{E}</math> Let <math>a = 2^x-4, b = 4^x-2</math>, so the equation becomes <math>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</math>. Hence <math>a=0 \implies 2^x=4 \implies x=2</math>, or <math>b=0 \implies 4^x=2 \implies x=\frac{1}{2}</math>, or <math>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</math>, but <math>2^x >0</math> for all <math>x</math>, so we cannot have <math>2^x = -3</math>; however, <math>2^x=2</math> works, giving <math>x=1.</math> Thus we have <math>3</math> solutions: <math>2, 1/2, 1</math>, whose sum is <math>\frac{7}{2}.</math>
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Note that <math>(2^x-4)+(4^x-2)=4^x+2^x-6,</math> so we let <math>a=2^x-4</math> and <math>b=4^x-2.</math> The original equation becomes <cmath>a^3+b^3=(a+b)^3.</cmath>
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We expand the right side, then rearrange:
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<cmath>\begin{align*}
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a^3+b^3 &= a^3+3a^2b+3ab^2+b^3 \\
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0 &= 3a^2b+3ab^2 \\
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0 &= 3ab(a+b).
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\end{align*}</cmath>
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<ul style="list-style-type:square;">
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  <li>If <math>a=0,</math> then <math>2^x-4=0,</math> from which <math>x=2.</math></li><p>
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  <li>If <math>b=0,</math> then <math>4^x-2=0,</math> from which <math>x=\frac12.</math></li><p>
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  <li>If <math>a+b=0,</math> then <math>4^x+2^x-6=0.</math> As <math>4^x=(2^x)^2,</math> we rewrite this equation, then factor:
 +
<cmath>\begin{align*}
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(2^x)^2+2^x-6&=0 \\
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(2^x-2)(2^x+3)&=0.
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\end{align*}</cmath>
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If <math>2^x-2=0,</math> then <math>x=1.</math> <p>
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If <math>2^x+3=0,</math> then there are no real solutions for <math>x,</math> as <math>2^x+3>3</math> holds for all real numbers <math>x.</math>
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</li><p>
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</ul>
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Together, the answer is <math>2+\frac12+1=\boxed{\textbf{(E) } \frac72}.</math>
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~Hapaxoromenon (Solution)
 +
 
 +
~MRENTHUSIASM (Reformatting)
  
 
== See also ==
 
== See also ==

Latest revision as of 03:33, 5 September 2021

Problem

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

$\textbf{(A) } \frac32 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } \frac52 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \frac72$

Solution

Note that $(2^x-4)+(4^x-2)=4^x+2^x-6,$ so we let $a=2^x-4$ and $b=4^x-2.$ The original equation becomes \[a^3+b^3=(a+b)^3.\] We expand the right side, then rearrange: \begin{align*} a^3+b^3 &= a^3+3a^2b+3ab^2+b^3 \\ 0 &= 3a^2b+3ab^2 \\ 0 &= 3ab(a+b). \end{align*}

  • If $a=0,$ then $2^x-4=0,$ from which $x=2.$

  • If $b=0,$ then $4^x-2=0,$ from which $x=\frac12.$

  • If $a+b=0,$ then $4^x+2^x-6=0.$ As $4^x=(2^x)^2,$ we rewrite this equation, then factor: \begin{align*} (2^x)^2+2^x-6&=0 \\ (2^x-2)(2^x+3)&=0. \end{align*} If $2^x-2=0,$ then $x=1.$

    If $2^x+3=0,$ then there are no real solutions for $x,$ as $2^x+3>3$ holds for all real numbers $x.$

Together, the answer is $2+\frac12+1=\boxed{\textbf{(E) } \frac72}.$

~Hapaxoromenon (Solution)

~MRENTHUSIASM (Reformatting)

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

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