Question

The set of values of ${}^{\prime }{x}^{\prime }$ satisfying the inequation, ${\log }_2(9-2^x)\le {10}^{{\log }_{10}(3-x)}$ are

• A. $x\in (-\infty ,0]$
  
• B. $x\in (-\infty ,1]$
• C. $x\in (-\infty ,0)$
• D. None of these

Answer 1

Answer: (A)

$x\in (-\infty ,0]$

Given,

${\log }_2(9-2^x)\le {10}^{{\log }_{10}3-x}$

$x<3$

${\log }_2(9-2^x)\le 3-x$

$9-2^x\le 8\times 2^{-x}$

$9\times 2^x-{{(2}^x)}^2\le 8$

${{(2}^x)}^2-8\cdot 2^x-2^x+8\ge 0$

$(2^x-8)(2^x-1)\ge 0$

$x\ge 3,x\le 0$

$\left[\right(x\ge 3)\cup (x\le 0\left)\right]\cap (x<3)$

$\Rightarrow x\le 0$