Question

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

• A. Two real solution
• B. No prime solution
• C. one integral solution
• D. No irrational solution

Answer 1

Answer: (B), (C), (D)

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