Question

How do you expand the logarithmic expression \[\ln \left( \dfrac{x}{3}y \right)\]?

Answer 1

Hint: For the question we are asked to find the logarithmic expansion of \[\ln \left( \dfrac{x}{3}y \right)\]. So, for the questions of these kind we will use the basic logarithmic formulae which are \[\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b\]and \[\ln \left( ab \right)=\ln a+\ln b\]. Using the above mentioned logarithmic formulae we will simplify the question and get the solution for the required question.

Complete step by step solution:
Firstly, for the question \[\ln \left( \dfrac{x}{3}y \right)\] we will use the basic logarithmic formula which is \[\ln \left( ab \right)=\ln a+\ln b\].
After using the formula we will simplify the equation. So, the equation will be reduced as follows.
\[\Rightarrow \ln \left( \dfrac{x}{3}y \right)\]
\[\Rightarrow \ln \left( \dfrac{x}{3} \right)+\ln \left( y \right)\]
Here after getting the above equation for the further simplification we will use the formulae \[\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b\] to the first term in the equation and we will keep the other term as it is.
So, after using the logarithmic formula \[\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b\] to the first term in the above equation we get the equation reduced as follows.
\[\Rightarrow \ln \left( \dfrac{x}{3} \right)+\ln \left( y \right)\]
\[\Rightarrow \ln \left( x \right)-\ln \left( 3 \right)+\ln y\]
Now, we will rearrange the equation which we got after all simplifications in the above to get the solution to look in a more familiar or an easier way.
So, after rearranging the equation will become as follows.
\[\Rightarrow \ln \left( x \right)+\ln y-\ln \left( 3 \right)\]
Therefore, the solution to the given question will be \[ \ln \left( x \right)+\ln y-\ln \left( 3 \right)\].

Note: We must be very careful in performing the calculations. We must have a very good knowledge in the concept of logarithms. We must know basic formulae like,
\[ \ln \left( \dfrac{a}{b} \right)=\ln a-\ln b\] and \[\Rightarrow \ln \left( ab \right)=\ln a+\ln b\]. We must not do mistake in using the formula for example for \[\ln \left( ab \right)\] if we use \[\ln a-\ln b\] as formula we get \[\ln \left( \dfrac{xy}{3} \right)=\ln \dfrac{x}{3}-\ln y\] which makes our whole solution wrong.