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

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

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

Answer

Explanation:

Step1: Apply log quotient rule

$\log_{12}\left(\frac{\frac{1}{2}}{8w}\right) = \log_{12}\frac{1}{2} - \log_{12}(8w)$

Step2: Apply log product rule

$\log_{12}(8w) = \log_{12}8 + \log_{12}w$

Step3: Substitute back to expression

$\log_{12}\frac{1}{2} - \left(\log_{12}8 + \log_{12}w\right)$

Answer:

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