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 $1$