Question 366565
If x=2 to the power y write 2 to the power 2y and 2 to the 2y+1 in terms of x and solve the equation 2 to the 2y+1 -5(2 to the y)+2=0


x = 2^y
2^(2y) is what in terms of x
2^(2y + 1) is what in terms of x
solve 2^(2y + 1) - 5(2^y) + 2 = 0


logarithmic rule: if b^y = x then logb x = y
in this case b = 2


if x = 2^y then log2 x = y


2^(2y) = 2^(2log2 x) = 2^(log2 x^2)
logarithmic rule: logb (m^n) = nlogb m


2^(2y + 1) = 2^(log2 x^2 + 1)
1 = log2 (2)
2^(2y + 1) = 2^(log2 x^2 + log2 2)
logarithmic rule: logb m + logb n = logb mn
2^(2y + 1) = 2^(log2 x^2 + log2 2) = 2^(log2 2x^2)


solve 2^(2y + 1) - 5(2^y) + 2 = 0
2^(log2 2x^2) - 5x + 2 = 0
2^(log2 2x^2) = 5x - 2
log2 (5x - 2) = log2 (2x^2)
if these 2 logs equal then 5x - 2 = 2x^2
-2x^2 + 5x - 2 = 0
2x^2 - 5x + 2 = 0
x^2 - (5/2)x + 1 = 0
(x - 2)(x - (1/2)) = 0, check with FOIL, First Outer Inner Last
x^2 - (1/2)x - 2x + 1 --> x^2 - (1/2)x - (4/2)x + 1
x^2 - (1/2)x - (4/2)x + 1 = x^2 - (5/2)x + 1, yes
x = 2 or x = 1/2
x = 2^y --> log2 x = y
log2 2 = 1
log2 (1/2) = -1