Question

Solve the following equations.
$x^2\cdot 2^{x+1}+2^{|x-3|+2}=x^2\cdot 2^{|x-3|+4}+2^{x-1}$

Answer 1

$x^2{.2}^{x+1}+2^{|x-3|+2}=x^2{.2}^{|x-3|+4}+2^{x-1}$

case $-Ix\ge 3$

$x^2{.2}^{x+1}+2^{x-3+2}=x^2{.2}^{x-3+4}+2^{x-1}$

$x^2{.2}^{x+1}+2^{x-1}=x^2{.2}^{x+1}+2^{x-1}$

$xϵ[3,\infty ]$

Case $-IIx<3$

$x^2{.2}^{x+1}+2^{-x+3+2}=x^2{.2}^{-x+3+4}+2^{x-1}$

$x^2.2{.2}^x+2^5{.2}^{-x}=x^2{.2}^{-x}{.2}^7+2^x.1/2$

$x^2.2{.2}^x-1/{2.2}^x=x^2{.2}^7{.2}^{-x}-2^5{.2}^{-x}$

$2^x(2x^2-1/2)=2^{-x}(2^7.x^2-2^5)$

$2^{2x}=\frac{2^5(2^2{x}^2-1)}{2x^2-1/2}=\frac{2^6(2^2.x^2-1)}{2^2{x}^2-1}$

if $x\ne 1/2$

$2^{2x}=2^6$

$2x=6$

$x=3$ No sol${}^n$

Ans $xϵ[3,\infty ]$