Question

values of $x$ for which the sixth term of the expansion of
$E={(3^{{\log }_3\sqrt{9^{|x-2|}}}+7^{\left(\frac{1}{5}\right){\log }_7\left[\right(4).3^{|x-2|}-9]})}^7$ is $567$, are

• A. $1$
• B. $2$
• C. $3$
• D. none of these

Answer 1

Answer: (A), (C)

$1$
$3$

Put $y=3^{{\log }_3\sqrt{9^{|x-2|}}}$ $\Rightarrow$ ${\log }_{3}y={\log }_3\sqrt{9^{|x-2|}}$
$\Rightarrow y=\sqrt{9^{|x-2|}}=3^{|x-2|}$
Now, put $z=7^{\left(\frac{1}{5}\right){\log }_7\left[\right(4).3^{|x-2|}-9]}$
${\log }_{7}z=\frac{1}{5}{\log }_7\left[\right(4).3^{|x-2|}-9]$
$={\left({\log }_7\left[\right(4).3^{|x-2|}-9]\right)}^{\frac{1}{5}}$
$\Rightarrow z={\left[\right(4).3^{|x-2|}-9]}^{\frac{1}{5}}$
Now, $E={(y+z)}^7$ and 6th term is given by
$t_6={{}^{7}C}_5{y}^{7-5}{z}^5=21{\left(3^{|x-2|}\right)}^2\left\{\right(4).3^{|x-2|}-9\}$
$\Rightarrow 567=21\left(3^{2|x-2|}\right)\left\{\right(4).3^{|x-2|}-9\}$
$\Rightarrow 27=\left[\right(4\left)3^{3|x-2|}\right]-\left[\right(9\left)\right(3^{2|x-2|}\left)\right]$
$\Rightarrow 27=\left(4\right)3^{3|x-2|}-\left(9\right)3^{2|x-2|}$
$\Rightarrow 4u^3-9u^2-27=0;u=3^{|x-2|}$
note that $u=3$ satisfies this equation
$3^{|x-2|}=3\Rightarrow |x-2|=1\Rightarrow x-2=\pm 1$
$x=2\pm 1=3or1$.