1 edit
Originally posted by battery123Start with the original expression and re-arrange:
2^(2Y+1)+7(2^Y)-15=0
find y to 2 d.p. and show your working
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