A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
Given, \[{{(a{{b}^{-1}})}^{2x-1}}={{(b{{a}^{-1}})}^{x-2}}\] \[\Rightarrow \] \[{{\left( \frac{a}{b} \right)}^{2x-1}}={{\left( \frac{b}{a} \right)}^{x-2}}\] \[\Rightarrow \] \[{{\left( \frac{a}{b} \right)}^{2x-1}}={{\left( \frac{a}{b} \right)}^{x-2}}=1\] \[\Rightarrow \] \[{{\left( \frac{a}{b} \right)}^{2x-1+x-2}}={{\left( \frac{a}{b} \right)}^{0}}\] On comparing, \[3x-3=0\] \[\Rightarrow \] \[x=1\]
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