Question

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

Answer 1

Answer: (A)

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 $