Question

Suppose that $x$ is real number such that $\frac{(27)\left(9^x\right)}{4^x}=\frac{3^x}{8^x}$, then the value of $(2{)}^{-\left(1+{\log }_{2}3\right)x}$, is

• A. 6
• B. 9
• C. 18
• D. 27

Answer 1

Answer: (D)

$(27)\cdot \frac{\left(9^x\right)}{3^x}=\frac{4^x}{8^x}\Rightarrow 27\cdot 3^x=\frac{1}{2^x}\Rightarrow 27=6^{-x}$ ....(1)
Now $(2{)}^{-\left(1+{\log }_{2}3\right)x}=(2{)}^{-x}\cdot (2{)}^{-{\log }_2(3{)}^x}=\frac{2^{-x}}{2^{{\log }_2{3}^x}}=\frac{2^{-x}}{3^x}=6^{-x}$ .....(2)
Hence $(2{)}^{-\left(1+{\log }_{2}3\right)x}=27$