WAEC - Mathematics (2015 - No. 7)
If 2n = y, Find 2\(^{(2 + \frac{n}{3})}\)
4y\(^\frac{1}{3}\)
4y\(^-3\)
2y\(^\frac{1}{3}\)
2y\(^-3\)
Explanation
If 2n = y,
then, 2\(^{(2 + \frac{n}{3})}\) = 22 x 2\(^\frac{n}{3}\)
= 4 x (2n)\(^{\frac{1}{3}}\)
But y = 2n, hence
2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\)
= 4y\(^\frac{1}{3}\)
then, 2\(^{(2 + \frac{n}{3})}\) = 22 x 2\(^\frac{n}{3}\)
= 4 x (2n)\(^{\frac{1}{3}}\)
But y = 2n, hence
2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\)
= 4y\(^\frac{1}{3}\)
Comments (0)
