(a) Given that ((sqrt{3} – 5sqrt{2})(sqrt{3} + sqrt{2}) = a + bsqrt{6}), find a and b.
(b) If (frac{2^{1 – y} times 2^{y – 1}}{2^{y + 2}} = 8^{2 – 3y}), find y.
Explanation
(a) ((sqrt{3} – 5sqrt{2})(sqrt{3} + sqrt{2}) = sqrt{3 times 3} + sqrt{3 times 2} – 5sqrt{2 times 3} – 5sqrt{2 times 2})
= (3 + sqrt{6} – 5sqrt{6} – 5(2))
= (-7 – 4sqrt{6})
(therefore) a = -7 and b = -4.
(b) (frac{2^{1 – y} times 2^{y – 1}}{2^{y + 2}} = 8^{2 – 3y})
(frac{2^{1 – y + y – 1}}{2^{y + 2}} = (2^{3})^{2 – 3y})
(2^{0 – (y + 2)} = 2^{6 – 9y} implies 2^{- y – 2} = 2^{6 – 9y})
(- y – 2 = 6 – 9y implies – y + 9y = 6 + 2)
(8y = 8 implies y = 1).