Question

If $ y={{2}^{\dfrac{1}{{{\log }_{x}}4}}} $ , then $ x $ is equal to
A. $ y $
B. $ {{y}^{2}} $
C. $ {{y}^{3}} $
D. none of these

Answer 1

Hint: We first take the given equation and try to simplify the indices value of logarithm. We use different identities like $ {{\log }_{m}}n=\dfrac{1}{{{\log }_{n}}m} $ , \[{{\log }_{{{m}^{p}}}}n=\dfrac{1}{p}{{\log }_{m}}n\], \[p{{\log }_{m}}n={{\log }_{m}}{{n}^{p}}\], $ {{a}^{{{\log }_{a}}m}}=m $ to simplify the expression. At the end we take the square to express $ x $ with respect to $ y $ .

Complete step-by-step answer:
We know that $ {{\log }_{m}}n=\dfrac{1}{{{\log }_{n}}m} $ .
We use this formula to simplify that
\[\dfrac{1}{{{\log }_{x}}4}={{\log }_{4}}x\].
So, $ y={{2}^{\dfrac{1}{{{\log }_{x}}4}}}={{2}^{{{\log }_{4}}x}} $ .
We know that \[{{\log }_{{{m}^{p}}}}n=\dfrac{1}{p}{{\log }_{m}}n\].
We use this formula to simplify that
\[{{\log }_{4}}x={{\log }_{{{2}^{2}}}}x=\dfrac{1}{2}{{\log }_{2}}x\].
We know that \[p{{\log }_{m}}n={{\log }_{m}}{{n}^{p}}\].
We use this formula to simplify that
\[\dfrac{1}{2}{{\log }_{2}}x={{\log }_{2}}{{x}^{\dfrac{1}{2}}}={{\log }_{2}}\sqrt{x}\].
So, $ y={{2}^{{{\log }_{4}}x}}={{2}^{{{\log }_{2}}\sqrt{x}}} $ .
We now use the identity theorem of $ {{a}^{{{\log }_{a}}m}}=m $ .
Therefore, $ y={{2}^{{{\log }_{2}}\sqrt{x}}}=\sqrt{x} $ .
Now we take the square of both sides of the equation to get the simplified solution.
We have $ {{y}^{2}}=x $ . The correct option is B.
So, the correct answer is “Option B”.

Note: In case the base is not mentioned then the general solution for the base for logarithm is 10. But the base of $ e $ is fixed for $ \ln $ . We also need to remember that for logarithm function there has to be a domain constraint. There are some particular rules that we follow in case of finding the condensed form of logarithm. We identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. Sometimes we also use 10 instead of $ e $ .