Question

The value of \[a = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 }}2\] then find the value of a
1). 1
2). 2
3). 3
4). 4
5). 5

Answer 1

Hint: We have to use different logarithmic properties to find the value of a. We will also rewrite 256 in terms of 4 and then 4 in terms of 2 to use the formula $\log {a^b} = b\log a$ and ${\log _a}a = 1$ to find the answer to this problem.

Complete step-by-step solution:
$a = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 }}2$
We can write 256 as $4^4$ and 2 as ${\sqrt 2 ^2}$. So, we will replace them in the above equation to convert them in the logarithmic formula. And we get,
$\Rightarrow a = {\log _2}{\log _2}{\log _4}{4^4} + 2{\log _{\sqrt 2 }}{\sqrt 2 ^2}$
Now, by using the formula $\log {a^b} = b\log a$. We will rewrite some terms of the above equation.
$\Rightarrow a = {\log _2}{\log _2}4{\log _4}4 + 2 \times 2{\log _{\sqrt 2 }}\sqrt 2 $
Now, we will use another formula ${\log _a}a = 1$, and rewrite some terms of the above equation as 1.
$\Rightarrow a = {\log _2}{\log _2}4 \times 1 + 4 \times 1$
$\Rightarrow a = {\log _2}{\log _2}4 + 4$
We can write 4 as 22. So,
$\Rightarrow a = {\log _2}{\log _2}{2^2} + 4$
Now, again we will use the formula $\log {a^b} = b\log a$ and rewrite some terms of this equation.
$\Rightarrow a = {\log _2}2{\log _2}2 + 4$
By using formula loga a = 1. We get,
$\Rightarrow a = {\log _2}2 \times 1 + 4$
Similarly, using these same formulas we will solve the equation further and find the answer.
$a = {\log _2}2 + 4$
$\Rightarrow a = 1 + 4$
$\Rightarrow a = 5$
The value of a is 5.
So, option (5) is the correct answer.

Note: This question consists of equations comprising logarithmic functions. So, we just need to use the appropriate logarithmic properties to solve the function and find the answer. In this case only 2 properties are used but students must remember all of them. Mistakes should be avoided in applying these logarithmic properties. $$