If 2ⁿ = y, Find 2 $^{(2 + \frac{n}{3})}$

4y $^\frac{1}{3}$

4y $^-3$

2y $^\frac{1}{3}$

2y $^-3$

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Correct answer is A

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 $^\frac{1}{3}$

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