Question

The values of $x$, for which the $6^{\text{th }}$ term in the expansion of ${\left(2^{{\log }_2\sqrt{\left(9^{x-1}+7\right)}}+\frac{1}{2^{(1/5){\log }_2\left(3^{x-1}+1\right)}}\right)}^7$ is $84$ are $x_1\&x_2$ then find value of $x_1+x_2$.

Answer 1

Answer: 3.00

${\left[2^{{\log }_2\sqrt{\left(9^{x-1}+7\right)}}+\frac{1}{2^{(1/5){\log }_2\left(3^{x-1}+1\right)}}\right]}^7$
$={\left[\sqrt{9^{x-1}+7}+\frac{1}{{\left(3^{x-1}+1\right)}^{1/5}}\right]}^7$
Here
$T_6=84$
$\Rightarrow 7C_5\cdot {\left(\sqrt{9^{x-1}+7}\right)}^2\cdot {\left(\frac{1}{{\left(3^{x-1}+1\right)}^{1/5}}\right)}^5=84$
$\Rightarrow \frac{21\cdot \left(9^{x-1}+7\right)}{\left(3^{x-1}+1\right)}=84$
$\Rightarrow 9^{x-1}+7=4\left(3^{x-1}+1\right)$
Put $3^{x-1}=t$
$\Rightarrow t^2+7=4t+4$
$\Rightarrow t^2-4t+3=0$
$\Rightarrow t=1$ or $t=3$