which expression is equivalent to $log_{12}left(\frac{\frac{1}{2}}{8w}\right)$?\n$log_{12}8…

which expression is equivalent to $log_{12}left(\frac{\frac{1}{2}}{8w}\right)$?\n$log_{12}8 - log_{12}\frac{1}{2}+log_{12}w$\n$log_{12}\frac{1}{2}-(log_{12}8+log_{12}w)$\n$log_{12}\frac{1}{2}-log_{12}8+log_{12}w$\n$log_{12}\frac{1}{2}+log_{12}8+log_{12}w$

which expression is equivalent to $log_{12}left(\frac{\frac{1}{2}}{8w}\right)$?\n$log_{12}8 - log_{12}\frac{1}{2}+log_{12}w$\n$log_{12}\frac{1}{2}-(log_{12}8+log_{12}w)$\n$log_{12}\frac{1}{2}-log_{12}8+log_{12}w$\n$log_{12}\frac{1}{2}+log_{12}8+log_{12}w$

Answer

Answer:

B. $\log_{12}\frac{1}{2}-(\log_{12}8 + \log_{12}w)$

Explanation:

Step1: Apply log - division rule

According to the logarithm division rule $\log_a\frac{M}{N}=\log_aM-\log_aN$, for $\log_{12}\frac{\frac{1}{2}}{8w}$, we have $\log_{12}\frac{1}{2}-\log_{12}(8w)$.

Step2: Apply log - multiplication rule

According to the logarithm multiplication rule $\log_a(MN)=\log_aM+\log_aN$, for $\log_{12}(8w)$, we get $\log_{12}(8w)=\log_{12}8+\log_{12}w$. So $\log_{12}\frac{\frac{1}{2}}{8w}=\log_{12}\frac{1}{2}-(\log_{12}8 + \log_{12}w)$.