A) \[\frac{1}{27}\]
B) \[\frac{1}{64}\]
C) \[\frac{1}{65}\]
D) None of these
Correct Answer: C
Solution :
[c] \[\frac{{{3}^{x}}}{1+{{3}^{x}}}=\frac{1}{9}\] \[\Rightarrow \] \[{{3}^{x}}.9=1+{{3}^{x}}\Rightarrow {{3}^{x}}(9-1)=1\] \[\Rightarrow \] \[{{3}^{x}}=\frac{1}{8}\] \[\Rightarrow \] \[{{9}^{x}}=\frac{1}{64}\] \[\therefore \frac{{{9}^{x}}}{1+{{9}^{x}}}=\frac{\frac{1}{64}}{1+\frac{1}{64}}=\,\,\,\,\,\,\,\,\frac{1/64}{65/64}=\frac{1}{65}\]
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