Question

Resolva as equações exponenciais

49^x - 8.7^x + 7 = 0 resposta {0,1}

Answer 1

$49^{x}-8.7^{x}+7=0 \\ \\ (7^{2})^{x}-8.7^{x}+7 = 0 \\ \\ 7^{2x}-8.7^{x}+7=0 \\ \\ y = 7^{x} \\ \\ y^{2}-8.y + 7 = 0 \\ \\ D = (-8)^{2}-4(1)(7) = 64-28 = 36 \\ \\ \sqrt{D} =6 \\ \\ y' = \frac{8+6}{2} = 7 \\ \\ y''=\frac{8-6}{2} =1 \\ \\ y = 7^{x} \\ \\ 7 = 7^{x} \\ \\ x = 1 \\ \\ 1 = 7^{x} \\ \\ 7^{0}=7^{x} \\ \\ x = 0 \\ \\ V=\{0,1\}$

Espero ter ajudado.

Answer 2

vamos lá...

$49^x-8.7^x+7=0 \\ \\ (7^2)^x-8.7^x+7=0 \\ \\ (7^x)^2-8.7^x+7=0 \\ \\ usamos~~7^x=y \\ \\ y^2-8y+7=0 \\ \\ \triangle=(-8)^2-4(7)=64-28=36 \\ \\ y'= \frac{8+6}{2} = \frac{14}{2} =7 \\ \\ y"= \frac{8-6}{2} = \frac{2}{2} =1$

$sendo \\ \\ 7^x=7~~~~~~~7^x=1 \\ 7^x=7^1~~~~~~7^x=7^0 \\ x=1~~~~~~~~~x=0$

S={0,1}