Question

The minimum value of $x$ satisfying the given inequality ${\log }_{10}(5\cdot 4^{x-1}+2x-20)\ge (1-x)\left({\log }_{10}\right(2.5)-1)$ is

Answer 1

${\log }_{10}(5\cdot 4^{x-1}+2x-20)\ge (1-x)\left({\log }_{10}\right(2.5)-1)\Rightarrow {\log }_{10}(5\cdot 4^{x-1}+2x-20)\ge (1-x)({\log }_{10}2.5-{\log }_{10}10)\Rightarrow {\log }_{10}(5\cdot 4^{x-1}+2x-20)\ge (1-x){\log }_{10}\left(\frac{1}{4}\right)\Rightarrow {\log }_{10}(5\cdot 4^{x-1}+2x-20)\ge {\log }_{10}{4}^{x-1}\Rightarrow 5\cdot 4^{x-1}+2x-20\ge 4^{x-1}\Rightarrow 4\cdot 4^{x-1}+2x-20\ge 0\Rightarrow 4^x+2x-20\ge 0$
$\therefore$ Minimum value of $x=2.$