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1. The value of $2 7^{- \frac{2}{3}}$ lies between = ?

A. 0 and 1

B. 1 and 2

C. 2 and 3

D. 3 and 4

Solution:

\begin{aligned} 2 7^{- \frac{2}{3}} \\ = {(3^{3})}^{- \frac{2}{3}} \\ = 3^{[3 \times (- \frac{2}{3})]} \\ = 3^{- 2} \\ = \frac{1}{3^{2}} \\ = \frac{1}{9} \\ Clearly, 0 < \frac{1}{9} < 1 \end{aligned}

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