If log _{a}(a b)=x, then log _{b}(a b) is | vedclass

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(D) Given $\log _{a}(a b)=x$.Using the property $\log(mn) = \log m + \log n$,we have:$\log _{a} a + \log _{a} b = x$Since $\log _{a} a = 1$,we get:$1

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$\frac{x}{x-1}$

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(D) Given $\log _{a}(a b)=x$.Using the property $\log(mn) = \log m + \log n$,we have:$\log _{a} a + \log _{a} b = x$Since $\log _{a} a = 1$,we get:$1 + \log _{a} b = x \Rightarrow \log _{a} b = x - 1$.Now,we need to find $\log _{b}(a b)$:$\log _{b}(a b) = \log _{b} a + \log _{b} b = \log _{b} a + 1$.Using the change of base property $\log _{b} a = \frac{1}{\log _{a} b}$,we substitute the value:$= \frac{1}{x - 1} + 1 = \frac{1 + (x - 1)}{x - 1} = \frac{x}{x - 1}$.

If $\log _{a}(a b)=x,$ then $\log _{b}(a b)$ is

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