nilai x yang memenuhi persamaan 2 log (x-2) + 2 log (x+4) = 4

Bab Logaritma
Matematika SMA Kelas X

^ = pangkat

²log (x - 2) + ²log (x + 4) = 4
²log ((x - 2) (x + 4)) = 4
x² + 4x - 2x - 8 = 2^4
x² + 2x - 8 - 16 = 0
x² + 2x - 24 = 0
(x + 6) (x - 4) = 0
x + 6 = 0
x = -6

x - 4 = 0
x = 4

syarat
x - 2 > 0
x > 2

x + 4 > 0
x > -4

Jadi, x = 4

✩✩ A ✩✩

[tex] {}^{2} log(x - 2) + {}^{2} log(x + 4) = 4 \\ {}^{2} log((x - 2)(x + 4)) = 4 \\ {}^{2} log( {x}^{2} + 2x - 8) = {}^{2} log( {2}^{4} ) \\ {x}^{2} + 2x - 8 = 16 \\ {x}^{2} + 2x - 8 - 16 = 0 \\ {x}^{2} + 2x - 24 = 0 \\ (x + 6)(x - 4) = 0 \\ x = - 6 \: \: atau \: \: x = 4[/tex]

masukkan x = 4
4-2 = positif, 4 + 4 = positif
masukkan x = -6
-6-2 = negatif, -6+4 =negatif
karena di uji coba tidak boleh negatif maka x = 4