3log (x + 2) = 1/3log (2x+3) ini gimana yaaa mohon sertakan cara juga yaaa trims

LogAritMa

3'log (x + 2) = (1/3)'log (2x + 3)
3'log (x + 2) = 3'log (2x + 3)^-1
x + 2 = 1/(2x + 3)
(x + 2)(2x + 3) = 1
2x² + 7x + 6 - 1 = 0
2x² + 7x + 5 = 0
(2x + 5)(x + 1) = 0
x = -5/2 atau x = -1

Nilai x yg memenuhi :
x = -1