Question

If $1+3{\log }_{10}\sqrt{2+x}+4{\log }_{10}\sqrt{2-x}=3{\log }_{10}\sqrt{4-x^2}$, then the value of $x$ is

• A. $\frac{199}{100}$
• B. $\frac{199}{200}$
• C. $\frac{99}{200}$
• D. $\frac{99}{100}$

Answer 1

The correct option is A $\frac{199}{100}$

$1+3{\log }_{10}\sqrt{2+x}+4{\log }_{10}\sqrt{2-x}=3{\log }_{10}\sqrt{4-x^2}$
For the $\log$ function to be defined, we get
$2+x>0x>-2...\left(1\right)2-x>0x<2...\left(2\right)4-x^2>0x^2<4\Rightarrow x\in (-2,2)...\left(3\right)$
From equation $\left(1\right)$,$\left(2\right)$ and $\left(3\right)$, we get
$x\in (-2,2)$

$1+3{\log }_{10}\sqrt{2+x}+4{\log }_{10}\sqrt{2-x}=3{\log }_{10}\sqrt{4-x^2}\Rightarrow 1+3{\log }_{10}\sqrt{2+x}+4{\log }_{10}\sqrt{2-x}=3{\log }_{10}\sqrt{2-x}+3{\log }_{10}\sqrt{2+x}\Rightarrow 1+{\log }_{10}\sqrt{2-x}=0\Rightarrow {\log }_{10}\sqrt{2-x}=-1\Rightarrow \sqrt{2-x}=\frac{1}{10}\Rightarrow 2-x=\frac{1}{100}\Rightarrow x=\frac{199}{100}$