Question

If $a,b>0,a,b\ne 1,c>0$, then ${\log }_{a}c=\frac{{\log }_{b}c}{{\log }_{b}a}=\left({\log }_{b}c\right)\left({\log }_{a}b\right)$

The solution set of ${\log }_3(3+\sqrt{x})+\frac{1}{2}{\log }_{\sqrt{3}}(1+x^2)=0$ will be

• A. $(1,2)$
• B. $(0,\infty )$
• C. $(2,3)$
• D. $\varphi$

Answer 1

Answer: (D)

$\varphi$

${\log }_3(3+\sqrt{x})+\frac{1}{2}{\log }_{\sqrt{3}}(1+x^2)=0$
$\Rightarrow {\log }_3(3+\sqrt{x})+{\log }_3(1+x^2)=0$ (1)
$[\because \frac{1}{m}{\log }_{a}b={\log }_{a^m}b]$
Above equation is valid when $x>0$
Clearly $\sqrt{x},x^2>0$
Therefore, $(3+\sqrt{x})(1+x^2)>3$
$\Rightarrow {\log }_3(3+\sqrt{x})(1+x^2)>1$
$\Rightarrow {\log }_3(3+\sqrt{x})+\frac{1}{2}{\log }_{\sqrt{3}}(1+x^2)>1$ ...[from (1)]
$\Rightarrow 0>1$
This is an ambiguous statement.
Therefore, $x=\left\{\varnothing \right\}$