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