Question

resolva a equação
a)$2^{x-3} =128$



c)$5^{x ^{2} -} ^{2x} =125$



d)$2^{x2 -3x-4^{} }= 2^{0}$

Answer 1

a)
lembrar que: $128=2^7$
Então:
$2^{x-3}=128\\ 2^{x-3}=2^7\\ x-3=7\\ x=7+3\\ x=10$

b)
Lembrar que: $125=5^3$

Então:
$5^{x^2-2x}=125\\ 5^{x^2-2x}=5^3\\ x^2-2x=3\\ x^2-2x-3=0\\ D=b^2-4.a.c\\ D=(-2)^2-4.1.(-3)=4+12=16\\ x_1=\frac{-b+ \sqrt{D} }{2.a}\\ x_1=\frac{-(-2)+ \sqrt{16} }{2.1}=\frac{2+4}{2}=\frac{6}{2}=3\\ x_2=\frac{-b- \sqrt{D} }{2.a}\\ x_2=\frac{-(-2)- \sqrt{16} }{2.1}=\frac{2-4}{2}=\frac{-2}{2}=-1$

c)
$2^{x^2-3x-4}=2^0\\ x^2-3x-4=0\\ D=b^2-4.a.c\\ D=(-3)^2-4.1.(-4)=9+16=25\\ x_1=\frac{-b+ \sqrt{D} }{2.a}\\ x_1=\frac{-(-3)+ \sqrt{25} }{2.1}=\frac{3+5}{2}=\frac{8}{2}=4\\ x_2=\frac{-b- \sqrt{D} }{2.a}\\ x_2=\frac{-(-3)- \sqrt{25} }{2.1}=\frac{3-5}{2}=\frac{-2}{2}=-1$