A) \[ab={{(c)}^{ab}}\]
B) \[{{a}^{b}}=c\]
C) \[ab=c(a+b)\]
D) None of these
Correct Answer: C
Solution :
\[{{(11)}^{a}}={{(19)}^{b}}={{(209)}^{c}}=z\] \[\Rightarrow \] \[11={{z}^{1/a}},\,19={{z}^{1/b}},\,\,209={{z}^{1/c}}\] Therefore, \[11\times 19=209\] \[{{z}^{1/a}}\times {{z}^{1/b}}={{z}^{1/c}}\] \[\Rightarrow \] \[{{Z}^{\left( \frac{1}{a}+\frac{1}{b} \right)=}}{{Z}^{\frac{1}{c}}}\] \[\Rightarrow \] \[\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\] \[\therefore \] \[c(a+b)=ab\]
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