Question

The value of the logarithmic function ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}16$ is equal to
(A) 0
(B) 1
(C) 2
(D) 4

Answer 1

Here in this question we have been asked to find the value of the given logarithmic function ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}16$ for answering that we will evaluate the given function from inner part firstly and then go for the outer one like firstly we will evaluate the value of ${{\log }_{2}}16$ and then ${{\log }_{2}}\left( {{\log }_{2}}16 \right)$ and so on using the logarithmic basic formula ${{\log }_{a}}{{a}^{x}}=x$ .

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of the given logarithmic function ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}16$ .
For answering that we will evaluate the given function from inner part firstly and then go for the outer one like firstly we will evaluate the value of ${{\log }_{2}}16$ and then ${{\log }_{2}}\left( {{\log }_{2}}16 \right)$ and so on.
From the basic concepts of logarithms we have learnt the logarithmic basic formula ${{\log }_{a}}{{a}^{x}}=x$ which we will use in between the steps.
So firstly we will evaluate the value of ${{\log }_{2}}16$ which will be given as $\Rightarrow {{\log }_{2}}16={{\log }_{2}}{{2}^{4}}\Rightarrow 4$ .
Now we will evaluate the value of ${{\log }_{2}}{{\log }_{2}}16$ which will result in
$\begin{aligned} & \Rightarrow {{\log }_{2}}{{\log }_{2}}16={{\log }_{2}}4 \\\ & \Rightarrow {{\log }_{2}}{{2}^{2}}=2 \\\ \end{aligned}$
Now we will evaluate the value of ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}16$ which will lead us to end up having the conclusion as
$\begin{aligned} & \Rightarrow {{\log }_{2}}{{\log }_{2}}{{\log }_{2}}16={{\log }_{2}}2 \\\ & \Rightarrow 1 \\\ \end{aligned}$
Hence we can conclude that the value of the given logarithmic function ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}16$ is equal to 1.
Therefore we will mark the option “B” as correct.

Note: While answering questions of this type we should be very sure with the concepts that we are going to apply and the calculations that we are going to perform in between the steps. Someone can make a calculation mistake and consider the solution as $\begin{aligned} & {{\log }_{2}}{{\log }_{2}}{{\log }_{2}}16={{\log }_{2}}{{\log }_{2}}4 \\\ & \Rightarrow {{\log }_{2}}4=2 \\\ \end{aligned}$
which will lead them to end up having a wrong conclusion.