Simplify the given expression: ${{\log }_{\sqrt{2}}}256$

Hint: First convert the term ${{\log }_{c}}d$ to the fraction of $\dfrac{\log d}{\log c}$ then change the number d, c in form of ${{a}^{b}}$ where a is the small possible prime factor of both c, d then make b according to it. Then do further calculation and get the results.

Complete step-by-step answer:
In the question we are asked to find the value of ${{\log }_{\sqrt{2}}}256$

For evaluating this we have to know this that if ${{\operatorname{loga}}^{b}}$ then it can be represented as,

$\dfrac{\log b}{\log a}$ .

So, we can represent ${{\log }_{\sqrt{2}}}256$ as

$\dfrac{{{\log }_{\sqrt{2}}}256}{\log \sqrt{2}}$ .

Now, we use the formula or the transformation rule which is $\log {{a}^{b}}$ which can be further written as b log a where b can be any real number whether rational or an irrational number.

Now we will represent each of the numerator and denominator of fraction $\dfrac{\log 256}{\log \sqrt{2}}$ in form of $\log {{a}^{b}}$such that their powers may vary but their base remain same.

So, we will represent both 256, $\sqrt{2}$ as in powers of 2.

So, we can represent the number 256 as in power of 2. So, the number 256 can be written as ${{2}^{8}}$ .

So, log 256 can also be written as $\log {{2}^{8}}$ . Hence using $\log {{a}^{b}}$

transformation which transforms into b log a we get, $\log {{2}^{8}}$ as 8 log 2.

Hence log 256 can be written as 8 log 2.

So, now we will have the $\sqrt{2}$ as in powers of 2. So, the $\sqrt{2}$ can be written as ${{2}^{\dfrac{1}{2}}}$ .

So, $\log \sqrt{2}$ can also be written as $\log {{2}^{\dfrac{1}{2}}}$. Hence using $\log {{a}^{b}}$ transformation which transforms into b log a we get, $\log {{2}^{\dfrac{1}{2}}}$ as $\dfrac{1}{2}\log 2$ .

Hence $\log \sqrt{2}$ can be written as $\dfrac{1}{2}\log 2$ .

Now we will write the fraction

$\dfrac{\log 265}{\log \sqrt{2}}$ in $\dfrac{8\log 2}{\dfrac{1}{2}\log 2}$.

Now by cancelling log 2 from both numerator and denominator and doing the calculations we get,

$\dfrac{\log 256}{\log \sqrt{2}}=\dfrac{8}{\dfrac{1}{2}}=8\times 2=16$

Hence the answer is 16.


Note: Students while doing these kind of problems while changing numbers into form ${{a}^{b}}$ they should take a as the smallest possible prime factor to avoid any problems in the solution in further steps and should also be careful about the calculation mistakes.