minta bantuannya guys

[tex]^{2}log[/tex] ([tex]x^{2}[/tex] - 2x + 1) = 0
[tex]^{2}log[/tex] ([tex]x^{2}[/tex] - 2x + 1) = [tex]^{2}log[/tex] 1
[tex]x^{2}[/tex] - 2x + 1 = 1
[tex]x^{2}[/tex] - 2x + 1 - 1 = 0
[tex]x^{2}[/tex] - 2x = 0
x ( x - 2) = 0
[tex]x_{1}[/tex] = 0 atau [tex]x_{2}[/tex] - 2 = 0
[tex]x_{1}[/tex] = 0
[tex]x_{2}[/tex] = 2

log ([tex]x^{2}[/tex] - 1) - log (x-1) = 1 + log (x - 8)
log [tex]\frac{x^{2}-1}{x-1}[/tex] = log 10 + log (x - 8)
log [tex]\frac{(x-1)(x+1)}{x-1}[/tex] = log 10(x - 8)
[tex]\frac{(x-1)(x+1)}{x-1}[/tex] = 10(x - 8)
x + 1 = 10x - 80
10x - x = 80 + 1
9x = 81
x = [tex]\frac{81}{9}[/tex] = 9

[tex]^{5}log^{2}[/tex] x - [tex]^{5}log[/tex] [tex]x^{6}[/tex] + 5 = 0
([tex]^{5}log[/tex]x)([tex]^{5}log[/tex]x) - 6[tex]^{5}log[/tex]x + 5 = 0
jika [tex]^{5}log[/tex]x = a, maka persamaan diatas menjadi:
(a x a) - 6a + 5 = 0
[tex]a^{2}[/tex] - 6a + 5 = 0
(a - 5)(a - 1) = 0
[tex]a_{1}[/tex] - 5 = 0 atau [tex]a_{2}[/tex] - 1 = 0
[tex]a_{1}[/tex] = 5
[tex]a_{2}[/tex] = 1
nilai a disubstitusikan ke persamaan [tex]^{5}log[/tex]x = a, maka
[tex]^{5}logx_{1}[/tex] = [tex]a_{1}[/tex]
[tex]^{5}logx_{1}[/tex] = 5
[tex]^{5}logx_{1}[/tex] = [tex]^{5}log5^{5}[/tex]
[tex]x_{1}[/tex] = [tex]5^{5}[/tex] = 3125
[tex]^{5}logx_{2}[/tex] = [tex]a_{2}[/tex]
[tex]^{5}logx_{2}[/tex] = 1
[tex]^{5}logx_{2}[/tex] = [tex]^{5}log[/tex]0
[tex]x_{2}[/tex] = 0
jadi, [tex]x_{1}[/tex] = 3125, [tex]x_{2}[/tex] = 0