Question

If ${\log }_4(3-x)+{\log }_{0.25}(3+x)={\log }_4(1-x)+{\log }_{0.25}(2x+1)$, it has

• A. only one real solution
• B. two real solutions
• C. no real solutions
• D. none of these

Answer 1

Answer: (A)

only one real solution

For given expression to be defined,
$3-x>0\Rightarrow x<3,1-x>0\Rightarrow x<1\Rightarrow \left(i\right)$
$3+x>0,2x+1>0\Rightarrow x>-1/2$
Thus domain is : $x\in (-\frac{1}{2},1)=D$ (say)
We have,
${\log }_4(3-x)+{\log }_{.25}(3+x)={\log }_4(1-x)+{\log }_{0.25}(2x+1)$
$\Rightarrow {\log }_4(3-x)+{\log }_{1/4}(3+x)={\log }_4(1-x)+{\log }_{1/4}(2x+1)$
$\Rightarrow {\log }_4(3-x)-{\log }_4(3+x)={\log }_4(1-x)-{\log }_4(2x+1),[{\log }_{1/a}b=-{\log }_{a}b]$
$\Rightarrow {\log }_4\frac{(3-x)}{(3+x)}={\log }_4\frac{(1-x)}{(2x+1)},[\because \log a-\log b=\log \frac{a}{b}]\Rightarrow \left(\frac{3-x}{3+x}\right)=\left(\frac{1-x}{2x+1}\right)\Rightarrow 6x+3-2x^2-x=3+x-3x-x^2\Rightarrow x^2-7x=0\Rightarrow x=0,7$
But $7\notin D,\therefore x=0$ is the only solution
Thus, it has only one real solutions.