Write the following expression as a single logarithm with coefficient 1. log₃(6c) + log₃1/12

Solution:

Given, \(log_{3}(6c)+log_{3}(\frac{1}{12})\)

We have to find a single logarithm with coefficient 1.

By property of logarithm,

\(log_{b}(\frac{x}{y})=log_{b}x-log_{b}y\)

\(log_{3}(\frac{1}{12})=log_{3}1-log_{3}12=0-log_{3}12=-log_{3}12\)

So, \(log_{3}(6c)+log_{3}(\frac{1}{12})=log_{3}(6c)-log_{3}(12)\)

Again by property of logarithm,

\(log_{b}x-log_{b}y=log_{b}(\frac{x}{y})\)

So, \(log_{3}(6c)-log_{3}(12)=log_{3}(\frac{6c}{12})\)

= \(log_{3}(\frac{c}{2})\)

Therefore, \(log_{3}(6c)-log_{3}(12)=log_{3}(\frac{c}{2})\).

Write the following expression as a single logarithm with coefficient 1. log₃(6c) + log₃1/12

summary:

The following expression log₃(6c) + log₃1/12 as a single logarithm with coefficient 1 is \(log_{3}(6c)-log_{3}(12)=log_{3}(\frac{c}{2})\).