log ab - log |b| = | vedclass

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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 $\frac{b}{|b|} = 	ext{sgn}(b)$,where $	ext{sgn}(b)$ is the sign of $b$,we have $\frac{ab}{|b|} = a \cdot 	ext{sgn}(b)$. However,for the expression $\log ab$ to be defined,$ab > 0$,which implies $a$ and $b$ must have the same sign. Thus,$\frac{ab}{|b|} = |a|$. Therefore,$\log ab - \log |b| = \log |a|$.

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

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