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# If$A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 }}2$, then,$A$ is equal to:A. 2B. 3C. 5D. 7

Last updated date: 17th Jun 2024
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Hint: The logarithmic function is the inverse function of the exponential function given by the formula${\log _b}a = c \Leftrightarrow {b^c} = \log a$, where b is the base of the logarithmic function. The logarithm is the mathematical operation that tells how many times a number or base is multiplied by itself to reach another number. There are five basic properties of the logarithm, namely Product rule, Quotient rule, Change of base rule, power rule, and equality rule.
The power rule of the logarithm is basically used to simplify the logarithm of power, rewriting it as the product of the exponent time to the logarithm base given by the formula${\log _a}{y^b} = b{\log _a}y$.
In this question, use the property of Power Rule of Logarithm to simplify the expression.

Complete step by step solution:As we can see in the given expression, we can see the base to the number 256 is 4, and for 2 the base is$\sqrt 2$; hence we try to write the number in terms of the base
We know
$256 = {2^8} = {4^4} \\ 2 = \sqrt 2 \times \sqrt 2 = {\left( {\sqrt 2 } \right)^2} \\$
Hence we can write the equation as
$A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 }}2 \\ = {\log _2}{\log _2}{\log _4}{\left( 4 \right)^4} + 2{\log _{\sqrt 2 }}{\left( {\sqrt 2 } \right)^2} \\$
Now by using the Power Rule of the logarithm, we can write
$A = {\log _2}{\log _2}{\log _4}{\left( 4 \right)^4} + 2{\log _{\sqrt 2 }}{\left( {\sqrt 2 } \right)^2} \\ = {\log _2}{\log _2}4 + 4 \\$[Since ${\log _a} a = 1$]
Now, again we can write $4 = {2^2}$
$A = {\log _2}{\log _2}4 + 4 \\ = {\log _2}{\log _2}{2^2} + 4 \\ = {\log _2}2 + 4 \\$[Again using power rule]
Now use the formula ${\log _a}a = 1$to solve
$A = {\log _2}2 + 4 \\ = 1 + 4 \\ = 5 \\$
Hence, the value of $A = 5$.

Note: Whenever log value is asked for a given number, try to bring the number in the form of the base of that logarithm by using the power rule. The logarithm of a base rule says${\log _a}a = 1$as the number and the base are the same.