For $x1$, if $(2 x)^{2 y}=4 e^{2 x-2 y}$, then $\left(1+\log _e 2 x\right)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

Question

For $x>1$, if $(2 x)^{2 y}=4 e^{2 x-2 y}$, then $\left(1+\log _e 2 x\right)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

Aicharya · Accepted Answer

$(2 x)^{2 y}=4 \cdot e^{2 x-2 y}$ $\Rightarrow 2 y \log 2 x=\log 4+2 x-2 y$ $\Rightarrow y=\frac{x+\log 2}{1+\log 2 x}$ $\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(1+\log 2 x)-(x+\log 2) \cdot \frac{1}{2 x} \cdot 2}{(1+\log 2 x)^2} \ & \Rightarrow(1+\log 2 x)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log 2 x-\log 2}{x}\end{aligned}$

For $x>1$, if $(2 x)^{2 y}=4 e^{2 x-2 y}$, then $\left(1+\log _e 2 x\right)^2 \frac{\mathrm{d}…

Solution