Question

Solve the following inequalities.
$8\cdot \frac{3^{x-2}}{3^x-2^x}>1+{\left(\frac{2}{3}\right)}^x$

Answer 1

$8.\frac{3^{x-2}}{3^x-2^x}>1+{\left(\frac{2}{3}\right)}^x$

$8.\frac{3^x{.3}^{-2}}{3^x-2^x}>1+{\left(\frac{2}{3}\right)}^x$ $[\because a^{m-n}=a^m.a^{-n}]$

$8\times \frac{3^x}{\frac{9}{3^x(1-{\left(\frac{2}{3}\right)}^x)}}>1+{\left(\frac{2}{3}\right)}^x$

$\frac{8}{9}>(1+{\left(\frac{2}{3}\right)}^x)(1-{\left(\frac{2}{3}\right)}^x)$ $[\because (a+b\left)\right(a-b)=a^2-b^2]$

$\frac{8}{9}>(1-{\left(\frac{2}{3}\right)}^{2x})$

${\left(\frac{2}{3}\right)}^{2x}>1-\frac{8}{9}$

${\left(\frac{4}{9}\right)}^x>\frac{1}{9}$

Apply log on both sides

$ln{\left(\frac{4}{9}\right)}^x>ln\left(\frac{1}{a}\right)$ $|\because ln{x}^n=nlnx|$

$x\left(ln\frac{4}{9}\right)>ln\left(\frac{1}{a}\right)$

$x>\frac{ln\left(1/a\right)}{ln\left(4/9\right)}\Rightarrow x>\frac{ln1-ln9}{ln4-ln9}$