Question

0,25^1-x + 0,5^-x-2 - 5(0,5)^1-x = 28


como resolvo??

Answer 1

Resposta:

x = log(2)14

Explicação passo-a-passo:

0,25^1-x + 0,5^-x-2 - 5(0,5)^1-x = 28

(25/100)^1-x + (5/10)^-x-2 - 5(5/10)^1-x = 28

(1/4)^1-x + (1/2)^-x-2 - 5(1/2)^1-x = 28

(1/2²)^1-x + (2-¹)^-x-2 - 5(2-¹)^1-x = 28

(2-²)^1-x + (2-¹)^-x-2 - 5(2-¹)^1-x = 28

(2)^(-2+2x) + (2)^(x+2) - 5(2)^(x-1) = 28

(2-²).2^(2x) + (2²).2^(x) - 5.2-¹(2)^(x) = 28

(1/4).2^(2x) + (4).2^(x) - 5.(1/2)(2)^(x) = 28

2^(2x) + (16).2^(x) - 5.(2)(2)^(x) = 28.4

2^(2x) + (16).2^(x) - 10(2)^(x) = 112

2^(2x) + 6.2^(x) = 112

2^(2x) + 6.2^(x) - 112 = 0

y=2^x

y² - 6y -112=0

y=(6±√484)/2

y=(6±22)/2

y' = 14

y'' = -8

raiz negativa descarta.

2^x = 14, aplica log de base 2 nos dois membros

log(2)2^x = log(2)14

xlog(2)2 = log(2)14

x.1 = log(2)14

x = log(2)14