If log (x+y) = log left( frac{3x - 3y}{2} rig | vedclass

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Question

(C) Given the equation: $\log (x+y) = \log \left( \frac{3x - 3y}{2} \right)$.By removing the logarithm from both sides,we get: $x + y = \frac{3x - 3y}

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$\log 5$

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(C) Given the equation: $\log (x+y) = \log \left( \frac{3x - 3y}{2} \right)$.By removing the logarithm from both sides,we get: $x + y = \frac{3x - 3y}{2}$.Multiplying both sides by $2$: $2x + 2y = 3x - 3y$.Rearranging the terms to solve for $x$ and $y$: $2y + 3y = 3x - 2x$.This simplifies to: $5y = x$,which implies $\frac{x}{y} = 5$.Using the logarithmic property $\log a - \log b = \log \left( \frac{a}{b} \right)$,we have: $\log x - \log y = \log \left( \frac{x}{y} \right)$.Substituting the value $\frac{x}{y} = 5$,we get: $\log x - \log y = \log 5$.

If $\log (x+y) = \log \left( \frac{3x - 3y}{2} \right)$,then $\log x - \log y =$

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