Question

The sixth term in the expansion of
$\left[2^{{\log }_2\sqrt{9^{x-1}+7}}+\frac{1}{2^{\frac{1}{5}{\log }_2\left(3^{x-1}+1\right)}}\right]$ is 84.
Then the number of values of $x$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2^{lo{g}_2^{\sqrt{9^{x-1}+7}}}=\sqrt{9^{x-1}+7}$
$2^{\frac{1}{2}lo{g}_2\left(3^{x-1}+1\right)}$
$={\left(3^{x-1}+1\right)}^{1/5}$
Also, the sixth term in the expansion is $84$
$\therefore {}^7{C}_5{\left(\sqrt{9^{x-1}+7}\right)}^2\cdot \frac{1}{3^{x-1}+1}=84$
$\Rightarrow \frac{9^{x-1}+7}{3^{x-1}+1}=\frac{84}{21}$
$\Rightarrow {\left(3^{x-1}\right)}^2-4\left(3^{x-1}\right)+3=0$
$\Rightarrow \left(3^{x-1}-1\right)\left(3^{x-1}-3\right)=0$
$\therefore 3^{x-1}=1$ or $3$
$\Rightarrow x=1$ or $2$.