For x1,if (2x)^{2y} = 4e^{2x-2y},then (1+log | vedclass

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

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

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(B) Given: $(2x)^{2y} = 4e^{2x-2y}$ Taking natural logarithm on both sides: $2y \log(2x) = \log 4 + 2x - 2y$ $2y \log(2x) = 2 \log 2 + 2x - 2y$ Dividing by $2$: $y \log(2x) = \log 2 + x - y$ $y(1 + \log 2x) = x + \log 2$ $y = \frac{x + \log 2}{1 + \log 2x}$ ... $(1)$ Now,differentiating $y \log(2x) = \log 2 + x - y$ with respect to $x$: $\frac{dy}{dx} \log(2x) + y \cdot \frac{1}{2x} \cdot 2 = 0 + 1 - \frac{dy}{dx}$ $\frac{dy}{dx} \log(2x) + \frac{y}{x} = 1 - \frac{dy}{dx}$ $\frac{dy}{dx} (1 + \log 2x) = 1 - \frac{y}{x} = \frac{x - y}{x}$ Substitute $y$ from $(1)$: $\frac{dy}{dx} (1 + \log 2x) = \frac{x - \frac{x + \log 2}{1 + \log 2x}}{x} = \frac{x(1 + \log 2x) - x - \log 2}{x(1 + \log 2x)}$ $\frac{dy}{dx} (1 + \log 2x) = \frac{x + x \log 2x - x - \log 2}{x(1 + \log 2x)} = \frac{x \log 2x - \log 2}{x(1 + \log 2x)}$ Multiplying both sides by $(1 + \log 2x)$: $(1 + \log 2x)^2 \frac{dy}{dx} = \frac{x \log 2x - \log 2}{x}$

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

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