If log (2 a-3 b)=log a-log b, then a= | vedclass

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Question

(A) Given the equation: $\log (2 a-3 b)=\log a-\log b$Using the logarithmic property $\log x - \log y = \log (x/y)$,we get:$\log (2 a-3 b)=\log (a/b)$

Vedclass · Accepted Answer

$\frac{3 b^{2}}{2 b-1}$

Vedclass · Answer

(A) Given the equation: $\log (2 a-3 b)=\log a-\log b$Using the logarithmic property $\log x - \log y = \log (x/y)$,we get:$\log (2 a-3 b)=\log (a/b)$By taking the antilog on both sides,we have:$2 a-3 b = a/b$Multiply both sides by $b$:$2 a b - 3 b^{2} = a$Rearrange the terms to isolate $a$:$2 a b - a = 3 b^{2}$$a(2 b-1) = 3 b^{2}$Therefore,$a = \frac{3 b^{2}}{2 b-1}$

If $\log (2 a-3 b)=\log a-\log b,$ then $a=$

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