Zadanie 1.32
a)
2^{x^2+2}=8
2^{x^2+2}=2^3
x^2+2=3 |-2
x^2=1
x=-1 lub x=1
b)
(\frac{2}{3})^{x^2-3x}=(\frac{9}{4})^{x-3}
(\frac{2}{3})^{x^2-3x}=(\frac{3}{2})^{2(x-3)}
(\frac{2}{3})^{x^2-3x}=(\frac{2}{3})^{-2(x-3)}
x^2-3x=-2(x-3)
x^2-3x=-2x+6
x^2-3x+2x-6=0
x^2-x-6=0
\Delta=b^2-4ac=(-1)^2-4*(-6)=1+24=25
\sqrt\Delta=5
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{-(-1)-5}{2}=\frac{-4}{2}=-2
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{1+5}{2}=3
c)
(\frac{5}{7})^{x^2-4x}=1,4^3
(\frac{5}{7})^{x^2-4x}=(\frac{14}{10})^3
(\frac{5}{7})^{x^2-4x}=(\frac{7}{5})^3
(\frac{5}{7})^{x^2-4x}=(\frac{5}{7})^{-3}
x^2-4x=-3
x^2-4x+3=0
\Delta=b^2-4ac=16-12=4
\sqrt\Delta=2
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{4-2}{2}=1
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{4+2}{2}=3