Question

помогите решить логарифмы
1 уравнение
2 неравенства

Answer 1

Уравнения:

$displaystyle (frac{1}{3})^{4-2x}=9\\3^{2x-4}=3^2\2x-4=2\x=3;\\\5^{x+2}+5^x=130\25cdot5^x+5^x=130\26cdot5^x=130\5^x=5\x=1;\\\3^{2x+1}-28cdot3^x+9=0\3cdot3^{2x}-28cdot3^x+9=0;big|;3^x=t\3t^2-28t+9=0\\frac{D}{4}:(frac{28}{2})^2-9*3=169=13^2\\t_1,_2=frac{14pm13}{3}; quad t_1=9, quad t_2=frac{1}{3};\\3^x=9 qquad quad 3^2=frac{1}{3}\3^x=3^2quad ; ; quad 3^x=3^{-1}\x=2; qquad quad x=-1.$


Неравенства:

$displaystyle 0,8^{2x-x^2} geq 1\\(frac{4}{5})^{2x-x^2} geq (frac{4}{5})^0\\2x-x^2 leq 0;big|;*(-1)\x(x-2) geq 0\\x_1=0; qquad x_2=2;\\+++0---2+++\\xin(-ifty;0] cup [2; +infty);\\\2^xcdot3^x textgreater 6^{2x^2}cdot frac{1}{6}\6^x textgreater 6^{-1}cdot 6^{2x^2}\6^x textgreater 6^{2x^2-1}\x textgreater 2x^2-1\2x^2-x-1 textless 0\D:1+8=9=3^2\\x_1,_2=frac{1pm3}{4}; qquad x_1=1, qquad x_2=-frac{1}{2};\\+++(-frac{1}{2})---1+++ \\xin (-frac{1}{2};1);$


$displaystyle 7^{x^2+4x} geq (2^x)^{x+4}\7^{x^2+4x} geq 2^{x^2+4x};big|;:2^{x^2+4x}\\(frac{7}{2})^{x^2+4x}=1\\(frac{7}{2})^{x^2+4x}=(frac{7}{2})^0\\x^2+4x geq 0\x(x+4) geq 0\\x_1=0; qquad x_2=-4;\\+++(-4)---0+++\\xin(-infty;-4)cup [0;+infty).$