Question

Показательные уравнения
1)$( sqrt[3]{5} ) ^{-3x+6} =25 ^{x+4}$
2)$6*25^x+5*36^x=11*30^x$
3)$6 ^{2x+1} = sqrt{6} ^{-4x+3}$
4)$7*49^x=50*7^x-7$

Answer 1

$(sqrt[3]{5})^{-3x+6}=25^{x+4}\ \(5^frac{1}{3})^{3(2-x)}=5^{2(x+4)}\ \5^{(2-x)}=5^{2(x+4)}\ \2-x=2(x+4)\ \2-x=2x+8\ \-x-2x=8-2\ \-3x=6\ \x=-2$

$6*25^x+5*36^x=11*30^x\ \6*5^{2x}+5*6^{2x}=11*5^x*6^x|/5^x*6^x\ \6*frac{5^x}{6^x}+5*frac{6^x}{5^x}=11\ \frac{5^x}{6^x}=t\ \6*t+5/t=11|*t\ \6t^2+5=11t\ \6t^2-11t+5=00\ \D=121-4*5*6=1\ \sqrt1=1\ \t_1=frac{11+1}{12}=1\ \t_2=frac{11-1}{12}=frac{10}{12}=frac{5}{6}\ \frac{5^x}{6^x}=1\ \= textgreater x_1=0\ \frac{5^x}{6^x}=frac{5}{6}\ \= textgreater x_2=1$
$6^{2x+1}=sqrt{6}^{-4x+3}\ 6^{2x+1}=6^{-2x+1.5}\ 2x+1=-2x+1.5\ 4x=0.5\ x=0.125$

$7*49^x=50*7^x-7 7*7^{2x}=50*7^x-7\ 7*7^{2x}-50*7^{x}=-7\ 7^x=t\ 7t^2-50t+7=0\ D=2500-4*49=2304\ sqrt{2304}=48\ t_1=(50-48)/14=1/7\ t_2=98/14=7\ 7^x=1/7\ x=-1\ 7^x=7\ x=1$

Answer 2

Спасибо!