Nilai x yang memenuhi ³ˣ⁺²log27 = ⁵log3

by

ialah

41
Pembahasan:
³ˣ⁺²log27 = ⁵log3
³ˣ⁺²log3³ = ⁵log3
³ˣ⁺²⅓log3 = ⁵log3
(3x + 2)⅓ = 5
3x + 2 = 5³
3x + 2 = 125
3x = 123
x = 41
  limit x mendekati 3 x² – 9/√x² + 16x – 5