Question

2(в степени х-1) + 2 (в степени 2х + 3) больше 17

Answer 1

$2^{x-1}+2^{2x+3}\ \textgreater \ 17\\\\2^{x}\cdot 2^{-1}+2^{2x}\cdot 2^3\ \textgreater \ 17\\\\ \dfrac{2^x}{2} +8\cdot2^{2x}-17\ \textgreater \ 0\\\\y:=2^x\ \textgreater \ 0\\\\ \dfrac{y}{2} +8y^2-17\ \textgreater \ 0\\\\16y^2+y-34\ \textgreater \ 0\\D=2177\\\\y_{1,2}= \dfrac{-1\pm \sqrt{2177} }{32} \\\\y\in(-\infty; \dfrac{-1- \sqrt{2177} }{32})\cup(\dfrac{-1+ \sqrt{2177} }{32};+\infty)$

т.к. y=2^x > 0
$y\in (\dfrac{-1+ \sqrt{2177} }{32};+\infty)$

$2^x\ \textgreater \ \dfrac{-1+ \sqrt{2177} }{32}\\\\x\ \textgreater \ \dfrac{\lg \dfrac{-1+ \sqrt{2177} }{32}}{\lg 2}$