Question

The equation ${|x+1|}^{{\log }_{(x+1)}(3+2x-x^2)}=(x-3)\left|x\right|$ has

• A. Unique solution
• B. two solutions
• C. no solutions
• D. more than two solutions

Answer 1

Answer: (C)

no solutions

${|x+1|}^{{\log }_{(x+1)}(3+2x-x^2)}=(x-3)\left|x\right|$

$x+1>0$ and $3+2x-x^2>0$

$\Rightarrow x>-1$ and $x^2-2x-3<0$

$\Rightarrow (x-3)(x+1)<0\Rightarrow xϵ(-1,3)$

$\therefore {\left|\right(x+1\left)\right|}^{{\log }_{(x+1)}(3+2x-x^2)}=(3+2x-x^2{)}^{{\log }_{(x+1)}(x+1)}$

$\Rightarrow 3+2x-x^2=(x-3)\left|x\right|$

for $x>0\to 3+2x-x^2=(x-3).x$

$\Rightarrow 3+2x-x^2=x^2-3x\Rightarrow 2x^2-5x-3=0$

$\Rightarrow (x-3)(2x+1)=0\Rightarrow x=3;x=\frac{-1}{2}$

But $xϵ(-1,3)$ So $x$ can't be 3

Also, $x>0$ for this case $\Rightarrow x$ can't be $\frac{-1}{2}$

for $x<0\to 3+2x-x^2=(x-3)(-x)$

$\Rightarrow 3+2x-x^2=3x-x^2\Rightarrow x=3$ (Not possible)

Hence, the equation has no solution.