A) \[x=0\]
B) \[x=1\]
C) \[x=2\]
D) \[x=-1\]
Correct Answer: B
Solution :
[b] \[\therefore {{3}^{x-}}^{1}+{{3}^{1-x}}=2\] \[\Rightarrow \frac{3x}{3}+\frac{3}{{{3}^{x}}}=2\] Let \[{{3}^{x}}=y\] \[\frac{y}{3}+\frac{3}{x}=2\] \[\Rightarrow {{y}^{2}}-6y+9=0\] \[{{(y-3)}^{2}}=0\] \[\Rightarrow y=3,3.\] Hence, \[{{3}^{x}}=y=3\] \[\Rightarrow x=1\]
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