Question

Resolva as equações :
A) log2(x+1)=4
B) log2(x^2+x+2)=3
C) log4(log2x)=1
D) logx+1(x^2+7)=2


(Me ajudem por favor)

Answer 1

Explicação passo-a-passo:

a)

$log_2(x + 1) = 4 \\\ C.E: x + 1 > 0 \\\ x > - 1 \\\ 2^4 = x + 1 \\\ x + 1 = 16 \\\ x = 15$

b)

$log_2(x^2 + x + 2) = 3 \\\ x^2 + x + 2 = 8 \\\ x^2 + x - 6 = 0 \\\ \Delta = 1^2 - 4.1.(- 6) \\\ \Delta = 2 + 24 \\\ \Delta = 25 \\\ x' = \frac{- 1 + 5}{2} \\\\ x' = 2 \\\ x" = \frac{- 1 - 5}{2} \\\\ x" = - 3 \\\ S = (2, - 3)$

c)

$log_4(log_2 x) = 1 \\\ log_2 x = 4 \\\ x = 2^4 \\\ x = 16$

d)

$log_{x + 1}(x^2 + 7) = 2 \\\ C.E.: x + 1 > 0 \\\ x > - 1 \\\ (x + 1)^2 = x^2 + 7 \\\ x^2 + 2x + 1 = x^2 + 7 \\\ 2x + 1 = 7 \\\ 2x = 6 \\\ x = 3$

Obs.: C.E. = condição de existência.

Espero ter ajudado.