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 $e$ .