Question

The number of values of $x\in R$ satisfying ${\log }_a(1-\sqrt{1+x})={\log }_{a^2}(3-\sqrt{1+x})$, where $a>1$, is

• A. $0$
• B. $1$
• C. $2$
• D. Infinitely many

Answer 1

The correct option is A $0$

Given : ${\log }_a(1-\sqrt{1+x})={\log }_{a^2}(3-\sqrt{1+x})$
For the $\log$ to be defined,
$1-\sqrt{1+x}>0,3-\sqrt{1+x}>0\Rightarrow \sqrt{1+x}<1,\sqrt{1+x}<3\Rightarrow 1+x<1\Rightarrow x\in (-\infty ,0)$
For the square root to be defined
$1+x\ge 0\Rightarrow x\ge -1\therefore x\in [-1,0)$

Now,
${\log }_a(1-\sqrt{1+x})=\frac{1}{2}{\log }_a(3-\sqrt{1+x})\Rightarrow {\log }_a(1-\sqrt{1+x}{)}^2={\log }_a(3-\sqrt{1+x})\Rightarrow (1-\sqrt{1+x}{)}^2=3-\sqrt{1+x}\Rightarrow 2+x-2\sqrt{1+x}=3-\sqrt{1+x}\Rightarrow x-1=\sqrt{1+x}$
Squaring both sides, we get
$\Rightarrow x^2-2x+1=1+x\Rightarrow x^2-3x=0\Rightarrow x=0,3$

But $x\in [-1,0)$, so the number of solution of the equation is $0$.