For x 1,if (2x)^{2y} = 4e^{2x-2y},then left | vedclass

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(B) Given the equation: $(2x)^{2y} = 4e^{2x-2y}$.Taking the natural logarithm $(\log_e)$ on both sides:$2y \log_e(2x) = \log_e 4 + (2x - 2y) \log_e e$

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$\frac{x \log_e 2x - \log_e 2}{x}$

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(B) Given the equation: $(2x)^{2y} = 4e^{2x-2y}$.Taking the natural logarithm $(\log_e)$ on both sides:$2y \log_e(2x) = \log_e 4 + (2x - 2y) \log_e e$Since $\log_e 4 = 2 \log_e 2$ and $\log_e e = 1$,we have:$2y \log_e(2x) = 2 \log_e 2 + 2x - 2y$Dividing by $2$:$y \log_e(2x) = \log_e 2 + x - y$Rearranging to solve for $y$:$y(1 + \log_e 2x) = x + \log_e 2$$y = \frac{x + \log_e 2}{1 + \log_e 2x} \quad \dots(i)$Differentiating both sides with respect to $x$:$\frac{dy}{dx} (1 + \log_e 2x) + y \left( \frac{1}{2x} \cdot 2 \right) = 1 - \frac{dy}{dx}$$\frac{dy}{dx} (1 + \log_e 2x) + \frac{y}{x} = 1 - \frac{dy}{dx}$$\frac{dy}{dx} (1 + \log_e 2x + 1) = 1 - \frac{y}{x}$$\frac{dy}{dx} (2 + \log_e 2x) = \frac{x - y}{x}$Wait,let us re-evaluate the derivative step:From $y(1 + \log_e 2x) = x + \log_e 2$,differentiate:$\frac{dy}{dx} (1 + \log_e 2x) + y \left( \frac{1}{x} \right) = 1$$\frac{dy}{dx} (1 + \log_e 2x) = 1 - \frac{y}{x} = \frac{x - y}{x}$Substitute $y$ from $(i)$:$\frac{dy}{dx} (1 + \log_e 2x) = \frac{x - \frac{x + \log_e 2}{1 + \log_e 2x}}{x} = \frac{x(1 + \log_e 2x) - (x + \log_e 2)}{x(1 + \log_e 2x)}$$= \frac{x + x \log_e 2x - x - \log_e 2}{x(1 + \log_e 2x)} = \frac{x \log_e 2x - \log_e 2}{x(1 + \log_e 2x)}$Multiplying both sides by $(1 + \log_e 2x)$:$(1 + \log_e 2x)^2 \frac{dy}{dx} = \frac{x \log_e 2x - \log_e 2}{x}$

For $x > 1$,if $(2x)^{2y} = 4e^{2x-2y}$,then $\left(1 + \log_e 2x\right)^2 \frac{dy}{dx}$ is equal to

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