Question

The value of $x$ for which the fourth term in the expansion of ${\left(5^{\frac{2}{5}{\left(log\right)}_5\sqrt{4^x+44}}+\frac{1}{5^{{\left(log\right)}_5\sqrt[3]{2^{x-1}+7}}}\right)}^8$ is $336$ can be equal to

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$T_4={\left(\_{}^{8}C\right)}_3{\left(5^{\frac{1}{5}{\left(log\right)}_5\left(4^x+44\right)}\right)}^5{\left(\frac{1}{5^{\frac{1}{3}{\left(log\right)}_5\left(2^{x-1}+7\right)}}\right)}^3$
$={\left(\_{}^{8}C\right)}_3\left(5^{{\left(log\right)}_5\left(4^x+44\right)}\right){\left(5^{{\left(log\right)}_5\left(2^{x-1}+7\right)}\right)}^{-1}$
$={\left(\_{}^{8}C\right)}_3\times \left(4^x+44\right)\times {\left(2^{x-1}+7\right)}^{-1}$
Given, $T_4=336$
$\Rightarrow \left(4^x+44\right){\left(2^{x-1}+7\right)}^{-1}=6$
$\Rightarrow 4^x+44=3\cdot 2^x+42$
$\Rightarrow {\left(2^x\right)}^2-3\cdot 2^x+2=0\Rightarrow 2^x=1,2$
$\Rightarrow x=0$ or $1$