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Algebra Level 3

3 x + 4 x + 1 2 x 1 6 x 9 x = 1 \large 3^x + 4^x + 12^x -16^x -9^x\, =\ 1 .

Find the sum of all real values of x \large x satisfying the above equation.


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The answer is 0.

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1 solution

Chew-Seong Cheong
Jul 15, 2016

Rearranging 3 x + 4 x + 1 2 x 1 6 x 9 x = 1 3^x+4^x+12^x - 16^x-9^x = 1 , we get 3 x + 4 x + 1 2 x = 1 6 x + 9 x + 1 3^x+4^x+12^x = 16^x+9^x+1 , where each terms of both the LHS and RHS is > 0 >0 . Therefore, we can apply AM-GM inequality.

  • LHS 3 x + 4 x + 1 2 x 3 14 4 x 3 = 3 \quad 3^x+4^x+12^x \ge 3\sqrt [3] {144^x} = 3 as equality occurs when 3 x = 4 x = 1 2 x = 1 3^x=4^x=12^x=1 or x = 0 x=0
  • RHS: 1 6 x + 9 x + 1 3 14 4 x 3 = 3 \quad 16^x+9^x+1 \ge 3\sqrt [3]{144^x} = 3 as equality occurs when 1 6 x = 9 x = 1 16^x=9^x=1 or x = 0 x=0 .

When 3 x + 4 x + 1 2 x 1 6 x 9 x = 1 3^x+4^x+12^x - 16^x-9^x = 1 , \implies LHS = = RHS, and LHS equals RHS only when they are minimum at x = 0 x=0 because for x < 0 x < 0 RHS is decreasing faster than LHS and for x > 0 x > 0 RHS is increasing faster then LHS.

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