• # question_answer The ninth term in the expansion of${{[{{3}^{{{\log }_{3}}\sqrt{{{25}^{x-1}}+7}}}+{{3}^{-\frac{1}{8}{{\log }_{3}}({{5}^{x-1}}+1)}}]}^{10}}$is equal  to 180, then $x$is equal to A)  1               B)         2                             C)         3                             D)         None of these

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 $y=1$ $\Rightarrow$               ${{5}^{x-1}}=1\Rightarrow {{5}^{x}}=5\Rightarrow x=1$