Question

How do you solve \[{{\log }_{2}}n=\dfrac{1}{3}{{\log }_{2}}27+{{\log }_{2}}36\]?

Answer 1

Hint: In this problem, we have to solve the given logarithmic expression, and find the value of n. We should use some logarithmic rules to solve this problem. We can see that one of the logarithmic terms has coefficient, we can get rid of it using the rule \[a\log n=\log {{n}^{a}}\] , we can then add the both logarithmic terms using the rule \[{{\log }_{x}}\left( n \right)+{{\log }_{x}}\left( m \right)={{\log }_{x}}\left( m\times n \right)\], we can then cancel the similar terms to get the value of n.

Complete step by step solution:
We know the given expression is,
 \[{{\log }_{2}}n=\dfrac{1}{3}{{\log }_{2}}27+{{\log }_{2}}36\]
We can now use some of the logarithmic rules to solve the given expression.
We can see that one of the logarithmic terms has coefficient, we can get rid of it using the rule
\[a\log n=\log {{n}^{a}}\]
We can now apply the above rule in the given expression, we get
\[\Rightarrow {{\log }_{2}}n={{\log }_{2}}{{\left( 27 \right)}^{\dfrac{1}{3}}}+{{\log }_{2}}36\]
We can now write the above step as,
\[\begin{align}
  & \Rightarrow {{\log }_{2}}n={{\log }_{2}}\sqrt[3]{27}+{{\log }_{2}}36 \\
 & \Rightarrow {{\log }_{2}}n={{\log }_{2}}3+{{\log }_{2}}36\text{ }\because {{3}^{3}}=9 \\
\end{align}\]
We can now add the above step using the rule,
\[{{\log }_{x}}\left( n \right)+{{\log }_{x}}\left( m \right)={{\log }_{x}}\left( m\times n \right)\]
We can now apply this rule, we get
\[\Rightarrow {{\log }_{2}}n={{\log }_{2}}\left( 3\times 36 \right)\]
We can now multiply the terms inside the brackets, we get
\[\Rightarrow {{\log }_{2}}n={{\log }_{2}}108\]
We can see that, in the above terms we have similar log terms, we can cancel them
\[\Rightarrow n=108\]
Therefore, the value of n = 108.

Note: Students make mistakes, while using the logarithmic rules to simplify the problem, we should always remember the logarithmic rules to solve these types of problems like \[{{\log }_{x}}\left( n \right)+{{\log }_{x}}\left( m \right)={{\log }_{x}}\left( m\times n \right)\] and \[a\log n=\log {{n}^{a}}\]. We should also know some perfect cube terms which are used in this problem.