A) \[a=b=2\]
B) \[~\text{a}=\text{b}=\text{1}\]
C) \[a=\frac{{{b}^{2}}}{1-b}\]
D) \[a=\frac{b}{b-1}\]
Correct Answer: D
Solution :
Since\[{{\log }_{10}}a+{{\log }_{10}}b={{\log }_{10}}(a+b)\] \[\therefore \] \[{{\log }_{10}}a\times b={{\log }_{10}}(a+b)\] \[\Rightarrow \] \[ab=a+b\] \[\Rightarrow \] \[ab-a=b\] \[\Rightarrow \] \[a(b-1)=b\] \[\Rightarrow \] \[a=\frac{b}{(b-1)}\]
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