Question

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 $\ln 3 = b$ ?

Answer 1

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 - $\ln \left( {\dfrac{1}{x}} \right) = - \ln x$