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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & (2 x)^{2 y}=4 e^{2 x-2 y} \Rightarrow 2 y \log (2 x)=\log 4+2 x-2 y \ & \Rightarrow 2 y \log (2 x)=2 \log 2+2 x-2 y\end{aligned}$ Different w.r.t $x$ $\begin{aligned} & \frac{d y}{d x} \cdot \log (2 x)+y \cdot \frac{1}{2 x} \cdot 2=0+1-\frac{d y}{d x} \ & \Rightarrow \frac{d y}{d x}(1+\log 2 x)=1-\frac{y}{x}=1-\frac{x+\log 2}{x(1+\log 2 x)} \ & {\left[\because 	ext { from }(1) y=\frac{x+\log 2}{(1+\log 2 x)}ight]} \ & \Rightarrow \frac{d y}{d x}(1+\log 2 x)^2=\frac{x(1+\log 2 x)-x-\log 2}{x} \ & \Rightarrow(1+\log 2 x)^2=\frac{d y}{d x}=\frac{x \log 2 x-\log 2}{x}\end{aligned}$

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

Solution