Question

The equation ${\log }_4(3-x)+{\log }_{0.25}(3+x)={\log }_4(1-x)+{\log }_{0.25}(2x+1)$ has

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

Answer 1

Answer: (D)

none of these

${\log }_4(3-x)+{\log }_{0.25}(3+x)={\log }_4(1-x)+{\log }_{0.25}(2x+1)$
Above equation is valid when,
$3-x>0,3+x>0,1-x>0,2x+1>0$
$\Rightarrow x\in (-\frac{1}{2},1)$
${\log }_4(3-x)+{\log }_{0.25}(3+x)={\log }_4(1-x)+{\log }_{0.25}(2x+1)$

$\Rightarrow {\log }_4(3-x)-{\log }_4(3+x)={\log }_4(1-x)-{\log }_4(2x+1)[\because {\log }_{a^n}{b}^m=\frac{m}{n}{\log }_{a}b]$

$\Rightarrow {\log }_4\left(\frac{3-x}{3+x}\right)={\log }_4\left(\frac{1-x}{2x+1}\right)[\because \log a-\log b=\log \frac{a}{b}]$

$\Rightarrow \left(\frac{3-x}{3+x}\right)=\left(\frac{1-x}{2x+1}\right)$

$\Rightarrow x^2-7x=0$
$\Rightarrow x=0,7$
Since, $x\in (-\frac{1}{2},1)$
Therefore, $x=0$
Ans: D