A) \[\frac{1}{p}+\frac{1}{q}\]
B) \[\frac{1}{2p}+\frac{1}{q}\]
C) \[\frac{1}{p}+\frac{1}{2q}\]
D) \[\frac{1}{2p}+\frac{2}{2q}\]
Correct Answer: B
Solution :
We have, log .x- = p log a, 2 log x = q log b Let \[k={{\log }_{x}}\sqrt{ab},\] then 2k log x = log ab = log a + log b \[=\frac{1}{p}\log x+\frac{2}{q}\log x\] and since \[\log x\ne 0,2k=\frac{1}{p}+\frac{2}{q}\] \[\Rightarrow \]\[k=\frac{1}{2p}+\frac{1}{q}\]
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