How do you prove that \[\left( {\dfrac{1}{2}} \right)(\ln 2 - \ln 4 + \ln 3) = \left( {\dfrac{1}{2}} \right)\ln \left( {\dfrac{3}{2}} \right)\] in exponential form?

Hint: First we will convert this equation into the form ${\log _a}b$. Then we will evaluate all the required terms. Then we will apply the property. Here, we are using $\ln a + \ln b = \ln (a \times b)$ and $\ln a - \ln b = \ln \left( {\dfrac{a}{b}} \right)$
logarithmic property. The value of the logarithmic function $\ln e$ is $1$

Complete step-by-step answer:
We will first apply the logarithmic property to convert the equation to solvable form. Compare the given equation with formula and evaluate the values of the terms.
Hence, here the values are:
$
  a = 2 \\
  b = 4 \\
  c = 3 \;
 $
By using the property, $\ln a - \ln b = \dfrac{{\ln a}}{{\ln b}}$ we can write,
 \[
   = \left( {\dfrac{1}{2}} \right)(\ln 2 - \ln 4 + \ln 3) \\
   = \left( {\dfrac{1}{2}} \right)(\ln \dfrac{2}{4} + \ln 3) \\
   = \left( {\dfrac{1}{2}} \right)(\ln \dfrac{1}{2} + \ln 3) \;
 \]
Now we will apply the property $\ln a + \ln b = \ln (a \times b)$ to the term \[\left( {\dfrac{1}{2}} \right)(\ln \dfrac{1}{2} + \ln 3)\] .
 \[
   = \left( {\dfrac{1}{2}} \right)\left( {\ln \dfrac{1}{2} + \ln 3} \right) \\
   = \left( {\dfrac{1}{2}} \right)\ln \left( {\dfrac{1}{2} \times 3} \right) \\
   = \left( {\dfrac{1}{2}} \right)\left( {\ln \dfrac{3}{2}} \right) \;
 \]
As, LHS = RHS
Hence, proved.

Note: Remember the logarithmic property precisely which is $\ln a + \ln b = \ln (a \times b)$ and $\ln a - \ln b = \ln \left( {\dfrac{a}{b}} \right)$. While comparing the terms be cautious. After the application of property when you get the final answer, tress back the problem and see if it returns the same values. Evaluate the base and the argument carefully. Also, remember that ${\ln _e}e = 1$.