Question

Solve the following inequalities.
$\sqrt{9^x-3^x+2}>3^x-9$

Answer 1

$\sqrt{9^x-3^x+2}>3^x-9$

Squaring on both sides

$\Rightarrow [\sqrt{9^x-3^x+2}{]}^x>[3^x-9{]}^2$

$\Rightarrow 9^x-3^x+2>(3^x{)}^2+(a{)}^x-2\times 3^x\times 9$

$\Rightarrow 9^x-3^x+2>9^x+81-18\left(3^x\right)$

$\Rightarrow 18\left(3^x\right)-3^x>81-2$

$\Rightarrow 17\left(3^x\right)>79$

$\Rightarrow 3^x>\frac{79}{17}$

Apply $lo{g}_3$ on both sides

$\Rightarrow lo{g}_3{3}^x>lo{g}_3\left(\frac{79}{17}\right)$ $[\because lo{g}_b{a}^n=nlo{g}_b^a]$

$\Rightarrow xlo{g}_3^3>lo{g}_{3}79-lo{g}_3^{17}$ $[\because log\left(\frac{a}{b}\right)=loga-logb]$

$\Rightarrow x>lo{g}_3^{79}-lo{g}_3^{17}$ $[\because lo{g}_a^a=1]$