Solution :
\(log_a x\) = p \(\implies\) \(a^p\) = x \(\implies\) a = \(x^{1/p}\)
Similarly \(b^q\) = \(x^2\) \(\implies\) b = \(x^{2/q}\)
Now, \(log_x \sqrt{ab}\) = \(log_x \sqrt{x^{1/p}x^{2/q}}\) = \(log_x x^{({1\over p}+{2\over q}){1\over 2}}\) = \(1\over {2p}\) + \(1\over q\).
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