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Posers and Puzzles

Posers and Puzzles

  1. 07 Nov '07 17:41
    2^(2Y+1)+7(2^Y)-15=0
    find y to 2 d.p. and show your working
  2. Standard member PBE6
    Bananarama
    07 Nov '07 18:22 / 1 edit
    Originally posted by battery123
    2^(2Y+1)+7(2^Y)-15=0
    find y to 2 d.p. and show your working
    Start with the original expression and re-arrange:

    2^(2y+1) + 7(2^y) - 15 = 0

    (2^2y)(2^1) + 7(2^y) - 15 = 0

    2(2^y)^2 + 7(2^y) - 15 = 0

    Now let 2^y = x and we end up with something quite familiar:

    2x^2 + 7x - 15 = 0

    Solving this using the quadratic formula, we get:

    x = (-7 +/- 13)/4

    Therefore:

    x = 3/2 or x = -5

    However, since x = 2^y, and the exponential function is always positive, we must reject the negative solution. Therefore:

    x = 3/2 = 2^y

    Taking the log base 2 ("log2" ) of both sides we have:

    y = log2(3/2) = log2(3) - log2(2) = log2(3) - 1

    Evaluating this expression numerically to 2 decimal places we get:

    y = 0.58
  3. 07 Nov '07 19:57
    Originally posted by PBE6
    Evaluating this expression numerically to 2 decimal places we get:

    y = 0.58
    Looks sound to me.