lf the ninth term. in the expansion of ${\left[3^{{\log }_3\sqrt{{25}^{x-1}+7}}+3^{-1/8{\log }_3\left(5^{x-1}+1\right)}\right]}^{10}$ is equal to 180 and $x>1$ then x is equal to 

In $(A+B{)}^n,T_9{=}^n{C}_8{A}^{n-8}{B}^8=180$

We know that $a^{{\log }_{a}n}=n$

$\begin{array}{l}\therefore A=\sqrt{{25}^{x-1}+7}=\sqrt{5^{2(x-1)}+7}\\ B={\left[\left(5^{x-1}+1\right)\right]}^{-1/8},n=10\\ T_9{=}^{10}{C}_6{A}^2{B}^8=180\end{array}$

$\therefore {}^{10}{C}_8\left(5^{2(x-1)}+7\right){\left[\left(5^{x-1}+1\right)\right]}^{-1}=180$

Now put $5^{x-1}=t$

$\begin{array}{l}\therefore \frac{10.9}{1.2}\left(t^2+7\right)(t+1{)}^{-1}=180\\ \therefore t^2+7=4(t+1)\text{ or }{t}^2-4t+3=0\\ \text{ or }(t-3)(t-1)=0\therefore t=3,1\\ \therefore 5^{x-1}=3,1\text{ or }{5}^x=15,5\\ \therefore x={\log }_{5}15\text{ or }x=1\end{array}$

But $x>1\therefore x={\log }_{5}15$