Find the value of ${\log _{\sqrt 2 }}\left( {256} \right)$.

Hint: In order to find the value of ${\log _{\sqrt 2 }}\left( {256} \right)$ , name it some value. Then by using the formula or rules of logarithms, $x = {\log _b}a \Leftrightarrow {b^x} = a$ , we get a relation between $a,b, x$. Then, simplify the terms to get the value of $x$.

Complete step by step solution:
We are given a value ${\log _{\sqrt 2 }}\left( {256} \right)$.
Naming the above equation as $x$ , that gives:
$x = {\log _{\sqrt 2 }}\left( {256} \right)$
From the logarithmic functions and rules, we know that if $x = {\log _b}a$ then it can be written as $x = {\log _b}a \Leftrightarrow {b^x} = a$.
So, comparing $x = {\log _b}a$ and $x = {\log _{\sqrt 2 }}\left( {256} \right)$ , we get $a = 256$ and $b = \sqrt 2 $.
Substituting the value of $a,b$ in ${b^x} = a$, we get:
${\left( {\sqrt 2 } \right)^x} = 256$ ………….(1)
We need to solve the above equation in for $x$:
Writing the prime factors of $256$, and we get:
$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Since, they are all in one single factor, as there are eight number of two, so writing it in terms of power of the number, and we get:
$
  2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^8} \\
   \Rightarrow 256 = {2^8} \\
 $
Substituting this in equation (1):
$
  {\left( {\sqrt 2 } \right)^x} = 256 \\
   \Rightarrow {\left( {\sqrt 2 } \right)^x} = {2^8} \\
 $
Since, we can write ${\left( {\sqrt a } \right)^b} = {a^{\dfrac{b}{2}}}$ , as it’s a square root; so, writing the above equation as this, we get:
\[
  {\left( {\sqrt 2 } \right)^x} = {2^8} \\
   \Rightarrow {2^{\dfrac{x}{2}}} = {2^8} \\
 \]

We can see that the left-hand side and the right-hand side are equal and have a common base $2$, so since they are equal, their powers would also be equal.

From this comparing the powers of both the sides, we get:
$\dfrac{x}{2} = 8$
Multiplying both the sides by $2$ in order to remove the denominator from the left side and to get the value of $x$ . So, we get:
$
  \dfrac{x}{2} \times 2 = 8 \times 2 \\
   \Rightarrow x = 16 \\
 $
and, $x$ was equal to ${\log _{\sqrt 2 }}\left( {256} \right)$.

Therefore, the value of ${\log _{\sqrt 2 }}\left( {256} \right)$ is $16$.

Note:
> If instead of the base $2$ in \[{2^{\dfrac{x}{2}}} = {2^8}\] , the powers was equal and the base was not, then we cannot compare the bases to get the value, that would give an error.
> In $x = {\log _b}a$, $a$ is called argument and $b$ is called base.