Question

The solution set of ${\log }_{1/3}(2^{x+2}-4^x)\ge -2$, is

• A. $(-\infty ,2-\sqrt{13})$
• B. $(-\infty ,2+\sqrt{13})$
• C. $(-\infty ,2)$
• D. $R$

Answer 1

The correct option is C $(-\infty ,2)$

${\log }_{1/3}(2^{x+2}-4^x)$ is defined, if
$2^{x+2}-4^x>0$
$\Rightarrow 4\left(2^x\right)-(2^x{)}^2>0$
$\Rightarrow 2^x(4-2^x)>0$
$\Rightarrow 4-2^x>0\Rightarrow 2^x<4\Rightarrow x<2\Rightarrow x\in (-\infty ,2)...\left(1\right)$

We have ,
${\log }_{1/3}(2^{x+2}-4^x)\ge -2$
$\Rightarrow (2^{x+2}-4^x)\le {\left(\frac{1}{3}\right)}^{-2}$
$\Rightarrow 2^{x+2}-(2^x{)}^2\le 9$
$\Rightarrow 4\left(2^x\right)-(2^x{)}^2-9\le 0$
$\Rightarrow (2^x{)}^2-4(2^x)+9\ge 0$
$\Rightarrow (2^x-2{)}^2+5\ge 0$
Which is true for all $x\in R...\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$
$x\in (-\infty ,2).$