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²log(x+1) < ²log8
(x+1) < 8
x < 7
²log (x+1) < ²log 8
x+1 < 8
x < 8-1
x < 7
² log (x+1) terdefinisi jika x+1 > 0 --> x > -1
jadi solusinya adalah -1 < x < 7