Question

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

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

Answer 1

Answer: (A), (B)

$(1,10)$
$({10}^{-\frac{9}{2}},1)$

${\left({\log }_{10}\left(100x\right)\right)}^2+{\left({\log }_{10}\left(10x\right)\right)}^2+{\log }_{10}x\le 14$ where, $x>0$
$\Rightarrow {\left({\log }_{10}\left(100\right)+{\log }_{10}x\right)}^2+{({\log }_{10}\left(10\right)+{\log }_{10}x)}^2+{\log }_{10}x\le 14[\because \log (ab)=\log a+\log b]$
$\Rightarrow {\left(2+{\log }_{10}x\right)}^2+{(1+{\log }_{10}x)}^2+{\log }_{10}x\le 14[\because \log a^m=m\log a\text{\&}{\log }_{a}a=1]$
Writing $u={\log }_{10}x$, the given inequality can be written as
${(2+u)}^2+{(1+u)}^2+u\le 14$
$\Rightarrow 2u^2+7u-9\le 0$
$\Rightarrow (2u+9)(u-1)\le 0$
$\Rightarrow -\frac{9}{2}\le u\le 1$
$\Rightarrow -\frac{9}{2}\le {\log }_{10}x\le 1$
$\Rightarrow {10}^{-\frac{9}{2}}\le x\le 10$
Ans: A,B