Question

How do you use the property of logarithms to find the exact value of \[{{\log }_{2}}6.{{\log }_{6}}4\]?

Answer 1

Hint: In the above mentioned question we can clearly see that we can solve this question with just the basic logarithmic property. The property that we will use is, we will be converting the base value to ln as we can see that the function mentioned in the first log is the base of the other log(the other function of log which is multiplied)

Complete step by step answer:
In the above question we are going to first change \[{{\log }_{2}}\text{ and }{{\log }_{6}}\]to \[{{\log }_{e}}\] and we can do this by following the basic principle of logarithm to change the of a logarithm function there are some rules that should hold true are the value of the base should be greater positive and should not be equal to 1 and the logarithmic argument should also be positive so as to change the base of the logarithm. Now to change the base we will follow this procedure: \[{{\log }_{a}}b=\dfrac{{{\log }_{x}}b}{{{\log }_{x}}a}\]
In this above logarithmic function “a'' is the base of the function “b” is the logarithmic function value which needs to be solved with the help of logarithm and “x” is the new base value of the logarithm function. As in the above stated question we can see that the “b” value of the first function is equal to the “a'' value of the other log function so by using this property we will be able to cancel out two log functions. So with the help of the base change property and also with the help of power property of logarithm we will able to solve the question easily and can get a final result so on solving we get It as:
\[=\dfrac{{{\log }_{e}}\text{6}}{{{\log }_{e}}2}\times \dfrac{{{\log }_{e}}4}{{{\log }_{e}}6}\]
In this we can see that \[{{\log }_{e}}6\] will get canceled out and we also know that square of 2 is 4 so we will be using this and then applies power method of log and we will get:
\[\begin{align}
  & =\dfrac{{{\log }_{e}}{{2}^{2}}}{{{\log }_{e}}2} \\
 & =\dfrac{2{{\log }_{e}}2}{{{\log }_{e}}2} \\
\end{align}\]
Now again we will cancel out \[{{\log }_{e}}2\] and we will get the final value as 2
So we get the exponential form of the whole logarithmic equation to be 2.

Note:
In the above stated question we are able to see that how these basics of logarithmic function can come in handy for such types of questions and hence can solve these questions without taking a lot of time, so try to remember these basic logarithmic function which can help you do that.