1. Oblicz log₃ (2 + log₄ 0,25)
2. Zapisz warunek x∈ <-6,4> używając symbolu wartości bezwzględnej.
zad1.
log₄0,25=-1 ,bo 4^-1=1/4=0,25
log₃ (2 + log₄ 0,25) = log₃ (2 + (-1))=log₃ (2 -1) = log₃ 1 = 0 , bo 3^0=1
zad2
x∈ <-6,4>6+4=10 10/2=5 -6+5=-1|x-(-1)| < 5|x+1| <5
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zad1.
log₄0,25=-1 ,bo 4^-1=1/4=0,25
log₃ (2 + log₄ 0,25) = log₃ (2 + (-1))=log₃ (2 -1) = log₃ 1 = 0 , bo 3^0=1
zad2
x∈ <-6,4>
6+4=10 10/2=5 -6+5=-1
|x-(-1)| < 5
|x+1| <5