Given that \( x>0, y>0, \) show that \( \log_{a} (xy) = \log_{a} x + \log_{a} y \) .

\begin{align} Let \ \log_{a} x & = p \ and \ \log_{a} y = q \quad \quad \enclose{circle}{\color{black}{1}} \\ \\ Rewrite \ as \ indices \\ \\ x & =a^p \ and \ y= a^q \\ \\ Then \ by \ the \ laws \ of \ indices \\ \\ xy & = a^p \times a^q \\ \\ xy & = a^{p+q} \\ \\ Rewrite \ as \ logarithms \\ \\ \log_{a} (xy) & = \log_{a} (a^{p+q}) \\ \\ & = p+q \\ \\ Then \ substituting \ from \ \enclose{circle}{\color{black}{1}} \\ \\ \log_{a} (xy) & = \log_{a} x + \log_{a} y \quad as \ required \end{align}

Given that \( x>0, y>0, \) show that \( \log_a (\frac{x}{y}) = \log_{a} x - \log_{a} y \) .

\begin{align} Let \ \log_{a} x & = p \ and \ \log_{a} y = q \quad \quad \enclose{circle}{\color{black}{1}}\\ \\ Rewrite \ as \ indices \\ \\ x & =a^p \ and \ y= a^q \\ \\ Then \ by \ the \ laws \ of \ indices \\ \\ \frac{x}{y} & = \frac{a^p}{a^q} \\ \\ \frac{x}{y} & = a^{p-q} \\ \\ Rewrite \ as \ logarithms \\ \\ \log_{a} (\frac{x}{y}) & = \log_{a} (a^{p-q}) \\ \\ & = p-q \\ \\ Then \ substituting \ from \ \enclose{circle}{\color{black}{1}} \\ \\ \log_{a} (\frac{x}{y}) & = \log_{a} x - \log_{a} y \quad as \ required \end{align}

Given that \( x>0, y>0, \) show that \( log_{a} x^n = n log_{a} x \) .

\begin{align} Let \ \log_{a} x & = p \quad \quad \enclose{circle}{\color{black}{1}}\\ \\ Rewrite \ as \ indices \\ \\ x & =a^p \\ \\ Then \ by \ the \ laws \ of \ indices \\ \\ x^n & = (a^p)^n \\ \\ x^n & = a^{np} \\ \\ Rewrite \ as \ logarithms \\ \\ \log_{a} x^n & = \log_{a} (a^{np}) \\ \\ & = np \\ \\ Then \ substituting \ from \ \enclose{circle}{\color{black}{1}} \\ \\ \log_a x^n & = n\log_{a} x \quad as \ required \end{align}