Determine the value of \[{\log _{\sqrt 2 }}256\]

Hint: We will use the identity, if ${\log _a}b = x$, then ${b^x} = a$ and rewrite the given problem by comparing with it. We will prime-factorise 256 and will express it in powers of $\sqrt 2 $. Then, compare the equation to find the value of \[{\log _{\sqrt 2 }}256\]

Complete step by step Answer:

We have to determine the value of \[{\log _{\sqrt 2 }}256\]
We will first simplify the given expression using various properties of the log.
We know that if a number is written as${\log _a}b = x$, then $a$ is called the base. Here, base is $\sqrt 2 $
If ${\log _a}b = x$, then ${b^x} = a$
Comparing with given expression with ${\log _a}b = x$, we have $a = \sqrt 2 $, $b = 256$ and we have to find the value of $x$.
According to the property of mentioned above, we can rewrite the expression as,
$\begin{gathered}
  {\log _{\sqrt 2 }}256 = x \\
   \Rightarrow {\left( {\sqrt 2 } \right)^x} = 256 \\
\end{gathered} $
We have to express 256 in powers of $\sqrt 2 $ to compare the base of the above expression.
We can do this by prime factorizing 256.


Therefore, $256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$, which implies $256 = {2^8}$
As it is known that $\sqrt 2 = {2^{\dfrac{1}{2}}}$, we can rewrite the above expression in the form,
${\left( {\sqrt 2 } \right)^x} = {\left( {\sqrt 2 } \right)^{16}}$
On comparing the above equation, we get $x = 16$
Hence, the value of \[{\log _{\sqrt 2 }}256\] is 16.

Note: This question can alternatively done by first simplifying the expression as \[{\log _{\sqrt 2 }}{\left( {\sqrt 2 } \right)^{16}}\]. Then, using the property, if ${\log _a}{b^n} = n{\log _a}b$ to write the expression as \[{\log _{\sqrt 2 }}{\left( {\sqrt 2 } \right)^{16}} = 16{\log _{\sqrt 2 }}\left( {\sqrt 2 } \right)\]. Also, it is known that \[{\log _a}a = 1\], then the value of \[{\log _{\sqrt 2 }}{\left( {\sqrt 2 } \right)^{16}}\] is 16.