Question

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

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

Answer 1

Answer: (A), (C)

3

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