Question

Решите неравенство: 3^-2x< либо =7*3^-x

Answer 1

$3^{-2x} \leq 7*3^{-x} \\ 3^{-2x} - 7*3^{-x} \leq 0 \\ 3^{-x}=a \\ a^2-7a \leq 0 \\ a(a-7) \leq 0$

_____$+$____$0$_____$-$_____$7$_____$+$______

$0\leq a \leq 7 \\ 0 \leq 3^{-x} \leq 7 \\ \\ \left \{ {{3^{-x} \geq 0} \atop {3 ^{-x} \leq 7}} \right. ~~~~~~ \left \{ {{3^{-x}\ \textgreater \ 0} \atop {3^{-x} \leq 3^{log_37}}} \right. ~~~~~ \left \{ {{xER} \atop {-x \leq log_37~|*(-1)}} \right. ~~~~~x \geq -log_37$

ответ: $x$ ∈ $[-log_37;+$беск.$)$