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

log (2x-3) + log (x-3) = log [(2x-3)(x-3)]
                                 = log (2x²-9x+9)

log (2x²-9x+9) = log 7
        2x²-9x+9 = 7
        2x²-9x+2 = 0
 gunakan rumus abc, x = (-b ± √(b² - 4ac))/2a

LogAritMa

log (2x - 3) + log (x - 3) = log 7
(2x - 3)(x - 3) = 7
2x² - 9x + 9 - 7 = 0
2x² - 9x + 2 = 0
a = 2
b = -9
c = 2

rumus abc
x1.2 = (-b ± √(b² - 4ac))/2a
x1.2 = (9 ± √65)/4

syarat numerus : > 0

NiLai x yg memenuhi :
x = (9 + √65)/4