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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$y^2$

Vedclass · Answer

(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

Similar Questions

$\log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}} = ?$

Find the solution of the equation $\log_{7}(\log_{5}(\sqrt{x^2 + x + 5})) = 0$.

If $x$ is a positive real number different from $1$ such that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $AP$, then

If ${a^x} = b$,${b^y} = c$,and ${c^z} = a$,then the value of $xyz$ is:

If $3^x = 4^{x-1}$,then $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module