expand: $log_{20}\frac{m}{n^{5}p}$\n$log_{20}m + 5log_{20}n-log_{20}p$\n$log_{20}m - 5log_{20}n+log_{20}p$\n$…

Explanation:

Step1: Apply quotient - rule of logarithms

The quotient - rule states that $\log_a\frac{x}{y}=\log_a x-\log_a y$. So, $\log_{20}\frac{m}{n^{5}p}=\log_{20}m-\log_{20}(n^{5}p)$.

Step2: Apply product - rule of logarithms

The product - rule states that $\log_a(xy)=\log_a x+\log_a y$. So, $\log_{20}(n^{5}p)=\log_{20}n^{5}+\log_{20}p$.

Step3: Apply power - rule of logarithms

The power - rule states that $\log_a x^{b}=b\log_a x$. So, $\log_{20}n^{5}=5\log_{20}n$. Then $\log_{20}\frac{m}{n^{5}p}=\log_{20}m-(5\log_{20}n + \log_{20}p)=\log_{20}m-5\log_{20}n-\log_{20}p$.

Answer:

$\log_{20}m - 5\log_{20}n-\log_{20}p$ (the third option)