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a) log⅓ x² > 0
zał. x²>0
x∈R\{0}
log⅓ x² > log⅓ 1 poniewaz f malejaca zmiana znaku
x²<1
x²-1<0
x=1∨x=-1
x∈(-1,1)∧x∈R\{0}
x∈(-1,1)\{0}
b) log₃ x² < 2
zał. x²>0
x∈R\{0}
log₃ x² < log₃ 4
x²<4
x²-4<0
x=2vx=-2
x∈(-2,2)∧x∈R\{0}
x∈(-2,2)\{0}