07 Nov '07 17:41

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

find y to 2 d.p. and show your working

find y to 2 d.p. and show your working

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False berry07 Nov '07 18:221 edit

Start with the original expression and re-arrange:*Originally posted by battery123***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- Joined
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