Question

If $x\in R$ satisfies $\left({\log }_{10}\right(100x){)}^2+({\log }_{10}\left(10x\right){)}^2+{\log }_{10}x\le 14$, then the solution set contains the interval

• A. $(1,10]$
• B. $[{10}^{-9/2},1]$
• C. $(0,\infty )$
• D. $(-1,\infty )$

Answer 1

Answer: (A), (B)

The correct options are
A $(1,10]$
B $[{10}^{-9/2},1]$

$\left({\log }_{10}\right(100x){)}^2+({\log }_{10}\left(10x\right){)}^2+{\log }_{10}x\le 14$
$\Rightarrow ({\log }_{10}100+{\log }_{10}x){)}^2+\left({\log }_{10}\right(10)+{\log }_{10}x{)}^2+{\log }_{10}x\le 14,[\because \log a+\log b=\log \left(ab\right)]$
$\Rightarrow (2+{\log }_{10}x){)}^2+(1+{\log }_{10}x{)}^2+{\log }_{10}x\le 14$
Substitute ${\log }_{10}x=y$
$\Rightarrow (y+2{)}^2+(y+1{)}^2+y\le 14$
$\Rightarrow 2y^2+7y-9\le 0$
$\Rightarrow (y-1)(2y+9)\le 0$
$\Rightarrow y\in [-\frac{9}{2},1]$
$\Rightarrow x\in [{10}^{-\frac{9}{2}},10]$