Question

If the $9^{\text{th}}$ 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 $180$, then the value(s) of $x$ is/are

• A. $2$
• B. $1$
• C. ${\log }_{5}3$
• D. ${\log }_{5}15$

Answer 1

Answer: (B), (D)

${\log }_{5}15$

Given : ${(3^{{\log }_3\sqrt{{25}^{x-1}+7}}+3^{-\frac{1}{8}{\log }_3(5^{x-1}+1)})}^{10}$
As ${25}^{x-1}>0,5^{x-1}>0$, so $\log$ function is defined for $x\in R$
We know that $a^{{\log }_{a}N}=N$
So,
${(3^{{\log }_3\sqrt{{25}^{x-1}+7}}+3^{-\frac{1}{8}{\log }_3(5^{x-1}+1)})}^{10}={(\sqrt{{25}^{x-1}+7}+{(5^{x-1}+1)}^{-\frac{1}{8}})}^{10}$
Now,
$T_9=T_{8+1}=180\Rightarrow {}^{10}{C}_8{\left(\sqrt{{25}^{x-1}+7}\right)}^2\cdot {(5^{x-1}+1)}^{\left(-\frac{1\times 8}{8}\right)}=180\Rightarrow {}^{10}{C}_8\frac{({25}^{x-1}+7)}{(5^{x-1}+1)}=180\Rightarrow 45\cdot \frac{{25}^{x-1}+7}{5^{x-1}+1}=180$
Let $t=5^{x-1}$, then
$\Rightarrow \frac{t^2+7}{t+1}=4\Rightarrow t^2-4t+3=0(\because t+1\ne 0)\Rightarrow (t-3)(t-1)=0\Rightarrow t=1,3\Rightarrow 5^{x-1}=1,3\Rightarrow x-1=0,{\log }_{5}3\therefore x=1,{\log }_{5}15$