Question

Evaluate the expression ${{\log }_{2}}{{\log }_{2}}{{\log }_{4}}256+2{{\log }_{\sqrt{2}}}2$.

Answer 1

Hint: The given problem is related to logarithms. Express the argument as a power of the base. Then use the following properties:
(i). $\log {{a}^{m}}=m\log a$ .
(ii). ${{\log }_{{{a}^{x}}}}b=\dfrac{1}{x}{{\log }_{a}}b$
(iii). ${{\log }_{a}}a=1$

Complete step-by-step answer:

The given expression is ${{\log }_{2}}{{\log }_{2}}{{\log }_{4}}256+2{{\log }_{\sqrt{2}}}2$. Let us assume ${{\log }_{2}}{{\log }_{2}}{{\log }_{4}}256$ to be a and $2{{\log }_{\sqrt{2}}}2$ be b.

Let us solve these separately.
First, let us solve for a.
$a={{\log }_{2}}{{\log }_{2}}{{\log }_{4}}256$
$\Rightarrow a={{\log }_{2}}{{\log }_{2}}\left( {{\log }_{4}}256 \right)$
Now, we can express 256 as \[=256=4\times 4\times 4\times 4={{4}^{4}}\] .
\[\Rightarrow a={{\log }_{2}}{{\log }_{2}}\left( {{\log }_{4}}{{4}^{4}} \right)\]
Now, we know $\log {{a}^{m}}=m\log a$ . So, ${{\log }_{4}}{{4}^{4}}=4{{\log }_{4}}4$ . We also know ${{\log }_{a}}a=1$ . So, ${{\log }_{4}}4=1$.
$\Rightarrow {{\log }_{4}}{{4}^{4}}=4$
\[\Rightarrow a={{\log }_{2}}{{\log }_{2}}4\]
\[={{\log }_{2}}\left( {{\log }_{2}}4 \right)\ \]
\[={{\log }_{2}}\left( {{\log }_{2}}{{2}^{2}} \right)\]
\[={{\log }_{2}}2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \text{since}\ {{\log }_{a}}a=1 \right]\]
\[=1\]
\[\therefore a=1\]
Now, let us solve for b.
$\begin{align}
  & b=2{{\log }_{\left( \sqrt{2} \right)}}2 \\
 & =2{{\log }_{{{2}^{\dfrac{1}{2}}}}}2 \\
\end{align}$
Now, we know, ${{\log }_{{{a}^{x}}}}b=\dfrac{1}{x}{{\log }_{a}}b$ .
Using the above expression,
$\begin{align}
  & 2{{\log }_{{{2}^{\dfrac{1}{2}}}}}2=2\times \dfrac{1}{\left( \dfrac{1}{2} \right)}\times {{\log }_{2}}2 \\
 & =2\times 2\times 1 \\
 & =4 \\
 & \therefore b=4 \\
\end{align}$
Now, we are asked to find the value of ${{\log }_{2}}{{\log }_{2}}{{\log }_{4}}256+2{{\log }_{\sqrt{2}}}2$ , i.e. a + b. We know, a = 1 and b = 4.
$\begin{align}
  & \Rightarrow a+b=1+4 \\
 & =5 \\
\end{align}$
Therefore, the answer is 5.

Note: While using the properties, make sure to use the terms properly. For example: While evaluating b, we used the property ${{\log }_{{{a}^{x}}}}b=\dfrac{1}{x}{{\log }_{a}}b$ . On comparing with $2{{\log }_{{{2}^{\dfrac{1}{2}}}}}2$ , we get $x=\dfrac{1}{2}$ . Some students get confused and write x = 2, which is wrong and can lead to wrong answers.