Practice Test Discussion

${If 289 = 17}^{x \frac{1}{5}}, then x = ?$

[OBC (PO), 2009]

$\frac{2}{5}$

None of these

Explanation

$\begin{array}{l} Here, \\ \Rightarrow 17^{x \frac{1}{5}} = 289 \\ \Rightarrow 17^{x + \frac{1}{5}} = {(17)}^{2} [∵ 289 = 17^{2}] \\ \Rightarrow x + \frac{1}{5} = 2 \\ \Rightarrow x = 2 - \frac{1}{5} = \frac{9}{5} \end{array}$

Hence, Option E is correct.

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