Question

If the ninth term in the expansion of ${[}^{{\log }_3\sqrt{{25}^{x-1}+7}}+3^{-1/8{\log }_3{5}^{x-1}+1}{]}^{10}$is equal to 180 and x > 1 then show that x is equal to ${\log }_5$ 15

Answer 1

In $(A+B{)}^n,T_9={}^n{C}_8{A}^{n-8}{B}^8=180$
We know that $a^{lo{g}_{a}n}=n$

$\therefore \sqrt{{25}^{x-1}+7}=\sqrt{5^{2(x-1)}+7}$

$b={\left[\right(5^{x-1}+1\left)\right]}^{-1/8},n=10$

$T_9={}^{10}{C}_8{A}^2{B}^8=180$

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

Now put $5^{x-1}=t$

$\therefore \frac{10.96}{1.2}(t^2+7)(t+7{)}^{-1}=180$

$\therefore t^2+7=4(t+1)or{t}^2-4t+3=0$

or ( t - 3) (t - 1) = 0 $\therefore$ t = 3 , 1

$\therefore 5^{x-1}=3,1or{5}^x=15,5$

$\therefore x={\log }_{5}15$ or x = 1

But x > 1 $\therefore x={\log }_{5}15$