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Q.
If all possible solutions to the equation $\log _4(3-x)+\log _{0.25}(3+x)=\log _4(1-x)+\log _{0.25}(2 x+1)$ are found, there will be
Continuity and Differentiability
Solution:
$\log _4\left(\frac{3- x }{3+ x }\right)=\log _4\left(\frac{1- x }{2 x +1}\right) \Rightarrow \frac{3- x }{3+ x }=\frac{1- x }{2 x +1}$
$(3-x)(2 x+1)=(1-x)(3+x) $
$5 x-2 x^2+3=3-2 x-x^2$
$x^2-7 x=0 \Rightarrow x=0 \text { or } x=7$
Reject $x=7$; as domain $=x \in\left(\frac{-1}{2}, 1\right)$.
$\therefore$ Only solution is $x =0$