Question

If $y=2^{\frac{1}{{\log }_x\left(4\right)}}$, then $x$ is equal to

• A. $y$
• B. $y^2$
• C. $y^3$
• D. None of these

Answer 1

The correct option is B

$y^2$

Explanation for the correct option:

Find the value of $x$.

In the equation $y=2^{\frac{1}{{\log }_x\left(4\right)}}$ take logarithm on both sides:

$$\begin{gathered} \log \left(y\right)=\log \left(2^{\frac{1}{{\log }_x\left(4\right)}}\right) \\ \Rightarrow \log \left(y\right)=\frac{1}{{\log }_x\left(4\right)}\log \left(2\right)\left[\mathbf{log}\mathbf{\left(}{\mathbf{a}}^{\mathbf{m}}\mathbf{\right)}\mathbf{=}\mathit{m}\mathbf{log}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\right] \\ \Rightarrow {\log }_x\left(4\right)=\frac{\log \left(2\right)}{\log \left(y\right)} \end{gathered}$$

Now in the left hand using the change of base rule ${\log }_b\left(a\right)=\frac{{\log }_{10}\left(a\right)}{{\log }_{10}\left(b\right)}$, change its base to $10$.

$\begin{array}{rcl}\frac{\log \left(4\right)}{\log \left(x\right)}& =& \frac{\log \left(2\right)}{\log \left(y\right)}\\ & \Rightarrow & \frac{\log \left(2^2\right)}{\log \left(x\right)}=\frac{\log \left(2\right)}{\log \left(y\right)}\\ & \Rightarrow & \frac{2\log \left(2\right)}{\log \left(x\right)}=\frac{\log \left(2\right)}{\log \left(y\right)}\left[\mathbf{log}\mathbf{\left(}{\mathbf{a}}^{\mathbf{m}}\mathbf{\right)}\mathbf{=}\mathit{m}\mathbf{log}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\right]\\ & \Rightarrow & 2\log \left(2\right)\times \frac{\log \left(y\right)}{\log \left(2\right)}=\log \left(x\right)\\ & \Rightarrow & 2\log \left(y\right)=\log \left(x\right)\\ & \Rightarrow & \log \left(y^2\right)=\log x\left[\mathit{m}\mathbf{log}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{log}\mathbf{\left(}{\mathbf{a}}^{\mathbf{m}}\mathbf{\right)}\right]\\ & \Rightarrow & y^2=x\end{array}$

So, the value of $x$ is $y^2$.

Hence, the correct option is B.