Question

Как решается показательное уравнение? (2,4)

Answer 1

$2)\frac{0,4^{x-2,5} }{\sqrt{2,5}} >2,5*0,16^{x-2} \\\\\frac{(\frac{2}{5})^{x-2,5}}{(\frac{5}{2})^{0,5}}>\frac{5}{2}*((\frac{2}{5})^{2})^{x-2}\\\\(\frac{2}{5})^{x-2,5}*(\frac{2}{5})^{-0,5}>(\frac{2}{5})^{-1}*(\frac{2}{5})^{2x-4}\\\\(\frac{2}{5})^{x-2,5-0,5} >(\frac{2}{5})^{-1+2x-4}\\\\(\frac{2}{5})^{x-3}>(\frac{2}{5})^{2x-5}\\\\0<\frac{2}{5}<1\\\\x-3<2x-5\\\\x-2x<-5+3\\\\-x<-2\\\\x>2\\\\Otvet:\boxed{ x\in(2;\infty)}$

$2)\frac{(\sqrt{2})^{x-7}}{3^{x-7}}>\frac{4\sqrt{2}}{243}\\\\(\frac{\sqrt{2}}{3})^{x-7}>(\frac{\sqrt{2}}{3})^{5}\\\\0<\frac{\sqrt{2} }{3}<1\\\\x-7<5\\\\x<12\\\\Otvet:\boxed {x\in(-\infty;12)}$