A) \[-1\]
B) 0
C) 1
D) 2
Correct Answer: D
Solution :
\[1<{{2}^{(-r+{{3}^{-a}})}}<2\] |
\[a={{\log }_{3}}{{\log }_{3}}2\] \[\Rightarrow {{3}^{a}}={{\log }_{3}}2\] \[\Rightarrow {{3}^{-a}}={{\log }_{2}}3\] |
\[0<-r+{{3}^{-a}}<1\,\,\,{{\log }_{2}}3-r>0\] \[\Rightarrow {{\log }_{2}}3>r,\] |
\[lo{{g}_{2}}3<1+r\] |
\[1<{{\log }_{2}}3<2\Rightarrow r=1\] |
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