Question

Показательные уравнения и неравенства .Все кроме 1 а и б

Answer 1

1 в)
$16*8^{2+3x}=1 \ 2^4*2^{6+9x}=2^0 \2^{10+9x}=2^0 \ 10+9x=0 \ x=- frac{10}{9}$
1 г)
$5*4^x+23*10^x-10*25^x=0 \ 5*( frac{2}{5} )^{2x}+23*( frac{2}{5} )^x-10=0 \ (frac{2}{5} )^x =t \ 5t^2+23t-10=0 \ D =529+200=729=27^2 \\ left[begin{array}{c} t_1= frac{-23-27}{10}=-5\t_2= frac{-23+27}{10}= frac{2}{5}end{array}right \ \ left[begin{array}{c}(frac{2}{5} )^x=-5\(frac{2}{5} )^x= frac{2}{5}end{array}right$
x = 1

2 a)
$3^{2x-x^2} textless 9 \ 3^{2x-x^2} textless 3^2 \ 2x-x^2 textless 2 \x^2-2x+2 textgreater 0$
x ∈ (-∞; 1) ∪ (2; +∞)

2б)
$10^x-8*5^x leq 0 \ 5^x(2^x-8) leq 0 \ 2^x-8leq 0 \ 2^xleq 2^3 \ x leq 3$
x ∈ (-∞; 3]

4.
$left { {{2^x*3^y=12} atop {2^y*3^x=18}} right. \ \ left { {{2^x*3^y=12} atop {2^{x-y}*3^{y-x}= frac{2}{3} }} right. \ \ left { {{2^x*3^y=12} atop { (frac{2}{3} )^{x-y}= frac{2}{3} }} right. \ \ left { {{2^x*3^y=12} atop {x-y=1}} right.\ \ left { {{y=x-1} atop {2^x*3^{x-1}=12}} right.\ \ left { {{6^x=36} atop {y=x-1}} right.\ \ left { {{x=2} atop {y=1}} right.$