Question

If all possible solutions to the equation ${\log }_4(3-x)+{\log }_{0.25}(3+x)={\log }_4(1-x)+{\log }_{0.25}(2x+1)$ are found, there will be

• A. 2 positive solutions
• B. no prime solution
• C. 1 positive and 1 negative solution
• D. two integral solutions

Answer 1

Answer: (B)

${\log }_4\left(\frac{3-x}{3+x}\right)={\log }_4\left(\frac{1-x}{2x+1}\right)\Rightarrow \frac{3-x}{3+x}=\frac{1-x}{2x+1}$
$(3-x)(2x+1)=(1-x)(3+x)$
$5x-2x^2+3=3-2x-x^2$
$x^2-7x=0\Rightarrow x=0\text{ or }x=7$
Reject $x=7$; as domain $=x\in \left(\frac{-1}{2},1\right)$.
$\therefore$ Only solution is $x=0$