Question

Solve the following inequalities.
${\log }_{x+0.2}2<{\log }_{x}4.$

Answer 1

${\log }_{x+0.2}2<{\log }_{x}4$

$\Rightarrow \frac{\log 2}{\log x+0.2}<\frac{\log 4}{\log x}$

$\Rightarrow \log 2\log x<2\log 2\left(\log \right(x+0.2\left)\right)$

$\Rightarrow \frac{\log x}{2}<\log (x+0.2)$

$\Rightarrow x<(x+0.2{)}^2$ [Remove $\log$ from both sides]

$\Rightarrow x<x^2+(0.2{)}^2+0.4x$

$\Rightarrow x^2+0.04+0.4x-x>0$

$\Rightarrow x^2+\frac{1}{25}-\frac{3}{5}x>0$

$\Rightarrow 25x^2+1-15x>0$

$\Rightarrow (x-\frac{3+\sqrt{5}}{10})(x+\frac{3-\sqrt{5}}{10})>0$, this will be possible only if,

$x<\frac{3-\sqrt{5}}{10}$ and $x>\frac{3+\sqrt{5}}{10}$

This is the solution.