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

In the given question, we are required to find the value of logarithmic function ${\log _2}{\log _2}{\log _2}16$.
The logarithmic number is converted to the exponential number. The exponential number is defined as the number of times the number is multiplied by itself. We will find the value of the expression by converting it into exponential form and comparing the powers.

Complete step-by-step solution:
Let us assume the expression as y.
So, we have, ${\log _2}{\log _2}{\log _2}16 = y$.
Now, we will solve the expression layer by layer.
So, let us consider ${\log _2}16 = t$.
The given number is in the form of a logarithmic number and we have to convert it into exponential form. The equation is in the form ${\log _x}y = b$ . To convert it into exponential form it is written as $y = {x^b}$, where x is the base of the log function.
Considering the equation${\log _2}16 = t$, when we compare to the general form y is $16$ and x is $2$. Therefore, it is written as ${2^t} = 16$ in exponential form.
The number $16$ is factored as:
$16 = 2 \times 2 \times 2 \times 2 = {2^4}$
Therefore, the number $16$ is written in the exponential form as ${2^4}$.
The above equation is written as:
$\Rightarrow {2^t} = {2^4}$
Since the bases are the same, the powers can be compared. So, we have,
$\Rightarrow t = 4$
Therefore, the value of t is $4$.
Now, we substitute this value into the expression. So, we get,
${\log _2}{\log _2}{\log _2}16 = {\log _2}{\log _2}4$
Considering ${\log _2}4 = u$, when we compare to the general form y is $4$ and b is $2$. Therefore, it is written as ${u^2} = 4$ in exponential form.
The number $4$ is factored as:
$4 = 2 \times 2 = {2^2}$
Therefore, the number $4$ is written in the exponential form as ${2^2}$.
The above equation is written as:
$\Rightarrow {2^u} = {2^2}$
Since the bases are the same, the powers can be compared. So, we have,
$\Rightarrow u = 2$
Therefore, the value of u is $2$.
Now, we substitute this value into the expression. So, we get,
${\log _2}{\log _2}{\log _2}16 = {\log _2}2$
Again, considering ${\log _2}2 = f$. When we compare the general form y is $2$ and b is $2$. Therefore, it is written as ${2^f} = 2$ in exponential form.
Since the bases are the same, the powers can be compared. So, we have,
$\Rightarrow f = 1$
Therefore, the value of f is $1$.
Now, we substitute this value into the expression. So, we get,
${\log _2}{\log _2}{\log _2}16 = 1$
Hence, the value of ${\log _2}{\log _2}{\log _2}16$ is $1$. Therefore, the option (B) is the correct answer.

Note: To solve the logarithmic equation we need to convert the equation to the exponential form and by using the concept of factorisation we can determine the value of logarithm. The exponential form of a number is defined as the number of times the number is multiplied by itself. The general form of logarithmic equation is ${\log _x}y = b$ and it is converted to exponential form as $y = {x^b}$. Hence we obtain the result.