# If $$log_a x$$ = p and $$log_b {x^2}$$ = q then $$log_x \sqrt{ab}$$ is equal to

## 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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