• question_answer If $\mathbf{lo}{{\mathbf{g}}_{\mathbf{a}}}\mathbf{(ab)=x,}$ then $\mathbf{lo}{{\mathbf{g}}_{\mathbf{b}}}$(ab) is: A)  $\frac{1}{x}$                         B)  $\frac{x}{x+1}$C)  $\frac{x}{1-x}$                       D)  $\frac{x}{x-1}$

(d): $lo{{g}_{a}}(ab)=\Leftrightarrow \frac{logab}{\log a}=x\Leftrightarrow \frac{loga+logb}{\log a}\text{ }=x$ $\Leftrightarrow 1+\frac{\log b}{\log a}=x\Leftrightarrow \frac{\log b}{\log a}=x-1$ $\Leftrightarrow \frac{\log a}{\log b}=\frac{1}{x-1}\Leftrightarrow 1+\frac{\log a}{\log b}=1+\frac{1}{x-1}$ $\Leftrightarrow \frac{\log b}{\log b}+\frac{\log a}{\log b}=\frac{x}{x-1}\,\,\,\Leftrightarrow \frac{\log b+\log a}{\log b}=\frac{x}{x-1}$ $\Leftrightarrow \frac{\log (ab)}{\log b}=\frac{x}{x-1}\,\,\,\Leftrightarrow {{\log }_{b}}(ab)=\frac{x}{x-1}$