Question

If $ {{\log }_{4}}2+{{\log }_{4}}4+{{\log }_{4}}x+{{\log }_{4}}16=6 $ , then value of x is
A. 64
B. 4
C. 8
D. 32

Answer 1

Hint: We will use the properties of logarithm to solve this question. As the base of logarithm is same for all terms we will use the property \[{{\log }_{a}}m+{{\log }_{a}}n=\log \left( m\times n \right)\] . Then simplify the expression to get the desired answer.

We have been given an expression $ {{\log }_{4}}2+{{\log }_{4}}4+{{\log }_{4}}x+{{\log }_{4}}16=6 $ .
We have to find the value of x.
We know that if the base of the logarithm is not given in the question we assume that it is a common logarithm. The logarithm of base 10 is called the common logarithm. Logarithms of base e are called natural logarithms.
Now let us simplify the given expression.
We know that \[{{\log }_{a}}m+{{\log }_{a}}n=\log \left( m\times n \right)\]
Now, applying the property in the given expression we get
\[\begin{align}
  & \Rightarrow \left( {{\log }_{4}}2+{{\log }_{4}}4 \right)+{{\log }_{4}}x+{{\log }_{4}}16=6 \\
 & \Rightarrow {{\log }_{4}}\left( 2\times 4 \right)+{{\log }_{4}}x+{{\log }_{4}}16=6 \\
 & \Rightarrow {{\log }_{4}}8+{{\log }_{4}}x+{{\log }_{4}}16=6 \\
\end{align}\]
Now, again applying the property in the obtained equation we get
\[\begin{align}
  & \Rightarrow \left( {{\log }_{4}}8+{{\log }_{4}}16 \right)+{{\log }_{4}}x=6 \\
 & \Rightarrow {{\log }_{4}}\left( 8\times 16 \right)+{{\log }_{4}}x=6 \\
 & \Rightarrow {{\log }_{4}}128+{{\log }_{4}}x=6 \\
\end{align}\]
Now, again applying the property in the obtained equation we get
\[\begin{align}
  & \Rightarrow \left( {{\log }_{4}}128+{{\log }_{4}}x \right)=6 \\
 & \Rightarrow {{\log }_{4}}128\times x=6 \\
 & \Rightarrow {{\log }_{4}}128x=6 \\
\end{align}\]
Now, we know that
 $ \begin{align}
  & {{\log }_{a}}x=y \\
 & \Rightarrow {{a}^{y}}=x \\
\end{align} $
So we can write the above obtained equation as
 \[\begin{align}
  & \Rightarrow {{\log }_{4}}128x=6 \\
 & \Rightarrow 128x={{4}^{6}} \\
\end{align}\]
Now, we know that we can write $ {{4}^{6}} $ as $ 4\times 4\times 4\times 4\times 4\times 4 $ .
Substituting the value and simplifying further we will get
 \[\begin{align}
  & \Rightarrow 128x=4\times 4\times 4\times 4\times 4\times 4 \\
 & \Rightarrow x=\dfrac{4\times 4\times 4\times 4\times 4\times 4}{128} \\
 & \Rightarrow x=32 \\
\end{align}\]
So by solving the given expression $ {{\log }_{4}}2+{{\log }_{4}}4+{{\log }_{4}}x+{{\log }_{4}}16=6 $ we get the value of $ x=32 $ .
Option D is the correct answer.

Note:
 To solve such types of questions students must have the knowledge of properties and rules of the logarithm. As these are simple questions, so be careful while doing calculations as a single calculation mistake leads to the incorrect answer.