\binom{(32)^{\frac{2}{3}} \cdot(72)^{\frac{4}{3}} \cdot(27)^{\frac{1}{2}}}/{(12)^{\frac{1}{3}} \cdot(16)^{\frac{3}{2}} \cdot(9)^{\frac{1}{4}}}=2^\alpha \cdot 3^\beta \Rightarrow 4 \alpha+5 \beta=

TS ICET 26 May 2023 Shift 2 Solved Paper

Question : 76 of 200

Marks: +1, -0

\;{(32)^{{2}/{3}} ⋅(72)^{{4}/{3}} ⋅(27)^{{1}/{2}}}/{(12)^{\;{1}/{3}} ⋅(16)^{\;{3}/{2}} ⋅(9)^{\;{1}/{4}}}=2^α ⋅ 3^β ⇒ 4 α+5 β=

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