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Question Number 130699 by LYKA last updated on 28/Jan/21

Answered by MJS_new last updated on 28/Jan/21

z=ln (x−2y+3) x−2y+3>0 ⇔ y<(x/2)+(3/2)∧x∈R ⇒ z∈R

z=ln(x−2y+3)x−2y+3>0⇔y<x2+32∧x∈R⇒z∈R

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