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

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