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 ${\log }_{\sqrt{2}}x+2{\log }_{2}x+{\log }_{1/2}x=9$ is

• A. $\left(4\right)$
• B. $\left(8\right)$
• C. $(8,16)$
• D. $\varphi$

Answer 1

Answer: (B)

$\left(8\right)$

${\log }_{\sqrt{2}}x+2{\log }_{2}x+{\log }_{1/2}x=9$
above equation is valid when $x>0$
$\Rightarrow 2{\log }_{2}x+2{\log }_{2}x-{\log }_{2}x=9[\because {\log }_{a^n}{b}^m=\frac{m}{n}{\log }_{a}b]$
$\Rightarrow {\log }_{2}x=3$
$\Rightarrow x=2^3=8$