If log (x+y)=log (x y)+a,where a is a constan | vedclass

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(A) Given the equation: $\log (x+y) = \log (xy) + a$ Using logarithmic properties,we can rewrite this as: $\log (x+y) - \log (xy) = a$ $\log \left( \f

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$-4$

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(A) Given the equation: $\log (x+y) = \log (xy) + a$ Using logarithmic properties,we can rewrite this as: $\log (x+y) - \log (xy) = a$ $\log \left( \frac{x+y}{xy} \right) = a$ $\frac{x+y}{xy} = e^a$ $\frac{1}{y} + \frac{1}{x} = e^a$ Differentiating both sides with respect to $x$: $-\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{x^2} = 0$ $\frac{dy}{dx} = -\frac{y^2}{x^2}$ Substituting $x=2$ and $y=4$: $\frac{dy}{dx} = -\frac{4^2}{2^2} = -\frac{16}{4} = -4$

If $\log (x+y)=\log (x y)+a$,where $a$ is a constant,then $\frac{dy}{dx}$ at $x=2$ and $y=4$ is

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