Question

Alguém me ajuda pfvrrrr

Answer 1

Resposta:

7) a) x = 6

8) x' = 5 e x" = 4

Explicação passo-a-passo:

7)

a) log 9 = 2

(x-3)

log 9 = 2.log (x-3) → log base 10

log 9 = log (x-3)²

(x-3)² = 9

x-3 = √9

x-3 = 3

x = 3 + 3

x = 6

8)

log (x-2) + log (x-3) = 1 + log (2x-7)

2 2 2

log (x-2) + log (x-3) - log (2x-7) = 1 × log 2

log (x-2).(x-3) - log (2x-7) = log 2

log (x-2).(x-3) = log (2x-7) + log 2

log (x-2).(x-3) = log 2.(2x-7)

(x-2).(x-3) = 2.(2x-7)

x² - 3x - 2x + 6 = 4x - 14

x² - 5x + 6 = 4x - 14

x² - 5x - 4x + 6 + 14 = 0

x² - 9x + 20 = 0

x = [9 ± √(9² - 4.1.20)]/2

x = [9 ± √(81 - 80)]/2

x = (9 ± 1)/2

x' = (9 + 1)/2 = 10/2 = 5

x" = (9 - 1)/2 = 8/2 = 4