Math Olympiad | A Nice Exponential Problem | 90% Failed to solve!

VIJAY Maths
VIJAY Maths
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Published 2024-05-05
All Comments (9)
  • @timc5768
    After ' m^3 - m = 2sqrt(3)', we could substitute ' q(sqrt(3)) = m', then 3q^3 - q = 2, so q =1, and (q - 1) (3q^2 + 3q + 2) = 0, and as '3^2 - 4(3)(2) < 0, 'q =1' is only real sol'n. So 'm = (9)^x = (3) ^(2x) = sqrt(3) = 3^(1/2)', so '2x = 1/2', and 'x = 1/4'.
  • @JPTaquari
    X = 0.25 1) 9^32 - 9^X = 2 SR of 3 2) 9^X * ( 9^2X - ) = " 3) Square both sides: 81^X * ( 9^4X + 1 - 2 . 9^2X ) = 12 To give 12, I go straight to 3 * 4 = 12, ergo: 81^X = 3 * We know mentally that X = 1/4 = 0.25 Test: 9^3X - 9^X = 2 SR of 3 1.732 * 3 - 1.732 = 2 * 1.732 1.732 * 3 = 1.732 * 3 3 * SR of 3 = 3 * SR of 3 Bingo from Brazil!!!!!!!
  • @igorsidorin3585
    3 roots of 3 is the root of 3 plus the root of three plus the root of three!! But not the root of three multiplied by the root of three and multiplied by the root of trez. Therefore, your decision is based on a false premise. Complete nonsense