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Question

(B) Using the logarithmic property $\log x - \log y = \log \left( \frac{x}{y} \right)$,we have: $\log ab - \log |b| = \log \left( \frac{ab}{|b|} \righ

Vedclass · Accepted Answer

$\log |a|$

Vedclass · Answer

(B) Using the logarithmic property $\log x - \log y = \log \left( \frac{x}{y} ight)$,we have: $\log ab - \log |b| = \log \left( \frac{ab}{|b|} ight)$. Since $|b| = b$ if $b > 0$ and $|b| = -b$ if $b $\frac{ab}{|b|} = a \cdot \frac{b}{|b|} = a \cdot 	ext{sgn}(b)$. Thus,$\log \left( \frac{ab}{|b|} ight) = \log |a|$ is the standard simplification assuming the domain of the logarithm requires the argument to be positive. Specifically,$\log |ab| - \log |b| = \log \left| \frac{ab}{b} ight| = \log |a|$.

$\log ab - \log |b| = $

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