Question

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

Explanation:

Step1: Apply Quotient Rule

The quotient rule for logarithms states that $\log_b \frac{x}{y} = \log_b x - \log_b y$. So, for $\log_{20} \frac{m}{n^5 p}$, we can rewrite it as $\log_{20} m - \log_{20} (n^5 p)$.

Step2: Apply Product Rule

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

Step3: Apply Power Rule

The power rule for logarithms states that $\log_b x^k = k\log_b x$. So, $\log_{20} n^5 = 5\log_{20} n$.

Step4: Combine the Results

Substituting back, we have $\log_{20} m - (\log_{20} n^5 + \log_{20} p) = \log_{20} m - 5\log_{20} n - \log_{20} p$.

Answer:

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