16.3 Change of Bases

Let \(p=\log _{\displaystyle {a}}M\) and \(q=\log _{\displaystyle {a}}N\) where \(a>0\). Then \(M=a^{\displaystyle {p}}\) and \(N=a^{\displaystyle {q}}\).

\begin {align*} M\cdot N & = a^{\displaystyle {p}}\cdot a^{\displaystyle {q}}\\ & = a^{\displaystyle {p+q}} \end {align*}

\begin {align*} \log _{\displaystyle {a}}MN & = \log _{\displaystyle {a}}a^{\displaystyle {p+q}}\\ \log _{\displaystyle {a}}MN & =(p+q) \log _{\displaystyle {a}}a\\\\ \implies \log _{\displaystyle {a}}MN & = p+q\\ \implies \log _{\displaystyle {a}}MN & = \log _{\displaystyle {a}}M+\log _{\displaystyle {a}}N \end {align*}

Now, we want to know how change bases. Say, change \(\log _{\displaystyle {a}}b\) to base \(c\).
So, let \(y=\log _{\displaystyle {a}}b\iff a^{\displaystyle {y}}=b\).

To the exponential form \((a^{\displaystyle {y}}=b)\) take log to base \(c\), we have that \(\log _{\displaystyle {c}}a^{\displaystyle {y}} =\log _{\displaystyle {c}}b\) \[\implies y\log _{\displaystyle {c}}a=\log _{\displaystyle {c}}b\implies y=\frac {\log _{\displaystyle {c}}b}{\log _{\displaystyle {c}}a}\] Thus \[\log _{a}b=\frac {\log _{c}b}{\log _{c}a} .\]

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