Question
Given that $\log 2=a$ and $\log 3=b$, write $\log \sqrt{96}$ terms of $a$ and $b$.

Answer

$\log 2=a \text { and } \log 3=b$
$\log \sqrt{ } 96=\frac{1}{2} \log (96)$
$=\frac{1}{2} \log \left(2^5 \times 3\right)$
$=\frac{1}{2}\left(\log 2^5+\log 3\right) \ldots . .[\because \log m n=\log m+\log n]$
$=\frac{1}{2}(5 \log 2+\log 3) \ldots \ldots\left[\cdot\left[\log m^n=n \log m\right]\right.$
$=\frac{5 a+b}{2} $

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