(A) express 4^x in terms of y given that 2^x = y. we know that 2² = 4. so 4^x = 2^{2^x} = (2^x)²= y² .so whenever you have 2^{C^D} it doesnt matter which order you solve. you can do (2^C)^D or (2^D)^C (B) solve 8(4^x) – 9(2^x) + 1 = 0 we can know that; 4^x = y² and 2^x= y so, 8(4^x) – 9(2^x) + 1 = 8(y²) - 9y +1 = 0 what we have left is a quadratic equation to solve (Ay + B)(Cy + D). we know that B and D must either be, both +1 or both -1. 8 can be formed by 4x2 or 8x1 so we know either A or C are 4 and 2 or 8 and 1. Ay X D = 8y X 1 = 8y. Cy X B = y X 1 = y. 8y + y = 9y because it is -9y we know that B and D must be -1. Therefore the quadratic is = (8y - 1)(y - 1) = 0. but we are not finished yet. we have to solve the quadratic. 8y - 1 = 0. 8y = 1 (divide by 8) y = 1/8. 2^x = 1/8 a negetive power flips the fraction and we know that 2³ is 8. therefore x = -3 second part of quadratic; (y - 1) = 0. y = 1. 2^x =1 x = 0

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(A) express 4^x in terms of y given that 2^x = y. (B) solve 8(4^x ) – 9(2^x ) + 1 = 0