Question

The value of x for which the sixth term in the expansion of
${[2^{{\log }_2\sqrt{9^{x-1}+7}}+\frac{1}{2^{\frac{1}{5}{\log }_2(3^{x-1}+1)}}]}^7$ is equal to 84 is

• A. 4
• B. 1 or 2
• C. 0 or 1
• D. 3

Answer 1

The correct option is B 1 or 2

Here we have,

${[2^{{\log }_2\sqrt{9^{x-1}+7}}+\frac{1}{2^{\frac{1}{5}{\log }_2(x-1)+1}}]}^7={[\sqrt{9^{x-1}+7}+(3^{x-1}+1{)}^{(-1/5)}]}^7$

The sixth term in the expansion is given by,

$T_6={}^7{C}_5(\sqrt{9^{x-1}+7}{)}^2\cdot {\left\{\right(3^{x-1}+1{)}^{(-1/5)}\}}^5\Rightarrow 84=21\cdot (9^{x-1}+7)\cdot (3^{x-1}+1{)}^{-1}\Rightarrow \frac{84}{21}=\frac{9^{x-1}+7}{3^{x-1}+1}\Rightarrow 4\cdot (3^{x-1}+1)=(3^{x-1}{)}^2+7\Rightarrow (3^{x-1}{)}^2-4\times 3^{x-1}+3=0\Rightarrow (3^{x-1}-1)(3^{x-1}-3)=0$

$\Rightarrow 3^{x-1}=1$ or $3$

$\Rightarrow 3^{x-1}=3^0$ or $3^1$

$\Rightarrow x-1=0$ or $1$

$\Rightarrow x=1,2$

Hence, option (B) is correct.