Question

If x be real, then solve the following equations :
$4^{-|x-1|}-4\cdot 2^{-|x-2|}+2=0$

Answer 1

Let us put $2^{-|x-2|}=t$
Since the base 2 is + ive and exponential function is +ive . Also $t=\frac{1}{2^{|x-2|}}$ $\le$1
$\therefore t\le$ 1 ........$1$
The given equation can be written as
$t^2-4t+2=0\therefore t=2\pm \sqrt{2}$
But $t=2+\sqrt{2}$ being > 1 is rejected by (1)
$\therefore t=2^{-|x-2|}=2-\sqrt{2}$
$\therefore 2^{-|x-2|}=\frac{1}{2-\sqrt{2}}=\frac{2+\sqrt{2}}{2}$
$\therefore |x-2|={\log }_2\left[\frac{2+\sqrt{2}}{2}\right]$
$\therefore x=2\pm \log \left[\frac{2+\sqrt{2}}{2}\right]$