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Question-130699




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
$${z}=\mathrm{ln}\:\left({x}−\mathrm{2}{y}+\mathrm{3}\right) \\ $$$${x}−\mathrm{2}{y}+\mathrm{3}>\mathrm{0}\:\Leftrightarrow\:{y}<\frac{{x}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\wedge{x}\in\mathbb{R}\:\Rightarrow\:{z}\in\mathbb{R} \\ $$

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