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)

$2^{log_2^{\sqrt{9^{x-1}+7}}} = \sqrt{9^{x-1}+7}$
$2^{\frac{1}{2}log_{2}\left(3^{x-1}+1\right)}$
$= \left(3^{x-1}+1\right)^{1/5}$
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 $2$.