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  4. The equation log x+1(x-0 ⋅ 5)= log x-0.5(x+1) has

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Q. The equation $\log _{x+1}(x-0 \cdot 5)=\log _{x-0.5}(x+1)$ has

Continuity and Differentiability

Solution:

$\log _{x+1}\left(x-\frac{1}{2}\right)=\log _{x-\frac{1}{2}}(x+1)$
$\frac{\log \left(x-\frac{1}{2}\right)}{\log (x+1)}=\frac{\log (x+1)}{\log \left(x-\frac{1}{2}\right)} $
$\log ^2\left(x-\frac{1}{2}\right)-\log ^2(x+1)=0$
$\left(\log \left(x-\frac{1}{2}\right)+\log (x+1)\right)\left(\log \left(x-\frac{1}{2}\right)-\log (x+1)\right)=0$
$\log \left(x-\frac{1}{2}\right)+\log (x+1)=0 \log \left(x-\frac{1}{2}\right)-\log (x+1)=0$
$x-\frac{1}{2}=\frac{1}{x+1}, \,\,\,\,x-\frac{1}{2}=x+1 $
$(2 x-1)(x+1)=2 \,\,\,\, \text { no solution } $
$2 x^2+x-1=2 $
$2 x^2+x-3=0 $
$\Rightarrow(2 x+3)(x-1)=0$
$\Rightarrow x =\frac{-3}{2}$ (rejected), $x =1$