If y = 2^{1/log_x 4},then x is equal to | vedclass

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(C) Given $y = 2^{1/\log_x 4}$.Taking logarithm on both sides with base $e$,we get $\ln y = \frac{1}{\log_x 4} \ln 2$.Using the change of base formula

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$y^2$

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(C) Given $y = 2^{1/\log_x 4}$.Taking logarithm on both sides with base $e$,we get $\ln y = \frac{1}{\log_x 4} \ln 2$.Using the change of base formula $\log_x 4 = \frac{\ln 4}{\ln x}$,we have $\ln y = \frac{\ln 2}{\frac{\ln 4}{\ln x}} = \frac{\ln 2 \cdot \ln x}{\ln 4}$.Since $\ln 4 = \ln(2^2) = 2 \ln 2$,the equation becomes $\ln y = \frac{\ln 2 \cdot \ln x}{2 \ln 2}$.Simplifying,we get $\ln y = \frac{\ln x}{2}$.Multiplying by $2$,we have $2 \ln y = \ln x$.Using the property $n \ln a = \ln(a^n)$,we get $\ln(y^2) = \ln x$.Therefore,$x = y^2$.

If $y = 2^{1/\log_x 4}$,then $x$ is equal to

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