How do you use properties of logarithms to write $ \ln \left( {\dfrac{2}{3}} \right) $ in terms of $ a $ and $ b $ if $ \ln 2 = a $ and $ \ln 3 = b $ ?

Hint: In order to write $ \ln \left( {\dfrac{2}{3}} \right) $ in terms of $ a $ and $ b $ for the given condition we need to know about the basic properties of the logarithms. Compare the following equation with one of the properties that is $ \ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y $ , put the value of $ a $ and $ b $ in the place needed and get the value.

Complete step by step solution:
We are given with $ \ln \left( {\dfrac{2}{3}} \right) $ , $ \ln 2 = a $ and $ \ln 3 = b $ .
From the properties of logarithm, we know that $ \ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y $ . On comparing $ \ln \left( {\dfrac{x}{y}} \right) $ with $ \ln \left( {\dfrac{2}{3}} \right) $ , we can write it as:
  $ \ln \left( {\dfrac{2}{3}} \right) = \ln 2 - \ln 3 $
As we are given that $ \ln 2 = a $ and $ \ln 3 = b $ . So, on replacing $ \ln 2 $ with $ a $ and $ \ln 3 $ with $ b $ in the above equation, we get the relation:
 $ \ln \left( {\dfrac{2}{3}} \right) = a - b $
Therefore, by using properties of logarithms we can write $ \ln \left( {\dfrac{2}{3}} \right) $ in terms of $ a $ and $ b $ as:
 $ \ln \left( {\dfrac{2}{3}} \right) = a - b $ , for $ \ln 2 = a $ and $ \ln 3 = b $ .
So, the correct answer is “ $ \ln \left( {\dfrac{2}{3}} \right) = a - b $ ”.

Note: The product rule - $ \ln \left( {xy} \right) = \ln x + \ln y $
The Quotient Rule - $ \ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y $
Log of a power - $ \ln \left( {{x^y}} \right) = y\ln x $
Log of $ 1 $ - $ \ln \left( 1 \right) = 0 $
Log of $ e $ - $ \ln \left( e \right) = 1 $
Log of reciprocal - $ \ln \left( {\dfrac{1}{x}} \right) = - \ln x $