A) 1
B) 2
C) 3
D) None of these
Correct Answer: A
Solution :
We have, \[{{[{{3}^{{{\log }_{3}}\sqrt{{{25}^{x-1}}+7}}}+{{3}^{-\frac{1}{8}{{\log }_{3}}{{5}^{x-1}}+1}}]}^{10}}\] \[={{[\sqrt{{{25}^{x-1}}+7}+{{({{5}^{x-1}}+1)}^{-1/8}}]}^{10}}\] \[(\because \,{{a}^{{{\log }_{a}}x}}=x)\] Here, \[{{T}_{9}}=180\] \[\Rightarrow \]\[{{\,}^{10}}{{C}_{8}}{{(\sqrt{{{25}^{x-1}}+7})}^{10-8}}{{\{{{({{5}^{x-1}}+1)}^{-1/8}}\}}^{8}}=180\] \[\Rightarrow \]\[{{\,}^{10}}{{C}_{8}}({{25}^{x-1}}+7).\frac{1}{({{5}^{x-1}}+1)}=180\] \[\Rightarrow \] \[\frac{{{25}^{x-1}}+7}{{{5}^{x-1}}+1}=\frac{180}{45}=4\] Let \[y={{5}^{x-1}},\]then above equation becomes \[\frac{{{y}^{2}}+7}{y+1}=4\] \[\Rightarrow \] \[{{y}^{2}}+7-4y-4=0\] \[\Rightarrow \] \[{{y}^{2}}-4y+3=0\] \[\Rightarrow \] \[(y-3)(y-1)+0\] \[\Rightarrow \] \[y=3,\,y=1\] If \[y=3\] \[\Rightarrow \] \[{{5}^{x-1}}=3\Rightarrow {{5}^{x}}=15\Rightarrow x={{\log }_{5}}15\] If \[y=1\] \[\Rightarrow \] \[{{5}^{x-1}}=1\Rightarrow {{5}^{x}}=5\Rightarrow x=1\]
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