Question

The equation ${\log }_{x+1}(x-0.5)={\log }_{x-0.5}(x+1)$ has which type of solution(s)?

• A. two real equations
• B. no prime solution
• C. one integral solution
• D. no irrational solution

Answer 1

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

The correct options are
B no prime solution
C one integral solution
D no irrational solution

${\log }_{x+1}(x-0.5)={\log }_{x-0.5}(x+1)$
Above equation is valid when $x-0.5>0,x-0.5\ne 1,x+1>0,x+1\ne 1$
$\Rightarrow x\in (0.5,\infty )-\left\{1.5\right\}$
${\log }_{x+1}(x-0.5)={\log }_{x-0.5}(x+1)$

$\Rightarrow \frac{\log (x-0.5)}{\log (x+1)}=\frac{\log (x+1)}{\log (x-0.5)}\cdots \cdots [\because {\log }_{a}b=\frac{\log b}{\log a}]$
$\Rightarrow {\left[\log (x-0.5)\right]}^2={\left[\log (x+1)\right]}^2$
$\Rightarrow \log (x-0.5)=\pm \log (x+1)$

$\log (x-0.5)=\log (x+1)$ cannot be true as it gives $1=-0.5$ which is not true

So, $\log (x-0.5)=-\log (x+1)$

$\log (x-0.5)=\log \frac{1}{(x+1)}$

$\Rightarrow (x-0.5)(x+1)=1$
$\Rightarrow x=1,-1.5$
Since, $x>0.5$
Therefore $x=1$
Ans: B,C,D