Question

How do you express in terms of $\ln 2$ and $\ln 3$ given $\ln 3\sqrt{2}$?

Answer 1

Hint: From the given question we have to express $\ln 3\sqrt{2}$ in terms of $\ln 2$ and $\ln 3$.
By using the properties of the logarithms we can solve the given question.
For the solution of the given question we have to use the two properties of the logarithms.

Complete step by step answer:
As we have been already discussed above we have to use the two properties of logarithms to solve the given question.
First we have to apply the below property to get the equation simplified,
First property is, $\ln \left( ab \right)=\ln \left( a \right)+\ln \left( b \right)$
By using the above property we can get the below logarithmic equation,
Simplified equation by applying the above property is written below, $\ln \left( 3\sqrt{2} \right)=\ln \left( 3 \right)+\ln \left( \sqrt{2} \right)$
We can write the square root as in terms of power of $\dfrac{1}{2}$
By using this we can get the below equation, $\ln \left( 3\sqrt{2} \right)=\ln \left( 3 \right)+\ln \left( {{2}^{\dfrac{1}{2}}} \right)$
Now, we have to use the second property to get the further simplification.
The second property which we have to use for the further simplification is mentioned below, $\ln \left( {{a}^{x}} \right)=x\ln \left( a \right)$
By using the above property we can get the below simplified form of logarithmic equation, $\ln \left( 3\sqrt{2} \right)=\ln \left( 3 \right)+\dfrac{1}{2}\ln \left( 2 \right)$

Hence the expression of $\ln \left( 3\sqrt{2} \right)$ in terms of $\ln 2$ and $\ln 3$ is $\ln \left( 3\sqrt{2} \right)=\ln \left( 3 \right)+\dfrac{1}{2}\ln \left( 2 \right)$.

Note: We should be very careful while using logarithmic equations. We should be well aware of the properties of logarithmic properties. There are some logarithm formulae here ${{\log }_{b}}{{m}^{p}}=p{{\log }_{b}}m$ , ${{\log }_{b}}mn={{\log }_{b}}m+{{\log }_{b}}n$ , ${{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n$ , ${{\log }_{b}}b=1$, ${{\log }_{b}}{{b}^{p}}=p$ , ${{\log }_{b}}1=0$ and ${{b}^{{{\log }_{b}}p}}=p$ . We can use these formulae while solving questions of this type.