Question

Solve the following inequality:
$4^{x+1}-{16}^x<2{\log }_{4}8$

Answer 1

$4^{x+1}-{16}^x<2{log}_{4}8$

Let $4^x=a$

$4a-a^2<{log}_{4}644a-a^2<3a^2-4a+3>0a\in (-\infty ,1)\cup (3,\infty )$

$a=4^x$ cannot be negative.

$4^x\in (0,1)\cup (3,\infty )x\in (-\infty ,0)\cup ({log}_{4}3,\infty ).$