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Logarithm Question (Important)

Please help me for answered this question

Note by Fadlan Semeion
4 years, 3 months ago

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\(P=2X^2+(3-X)^2\)) Minimum of P is at \(X=1\) or \((x,y)=(2,4)\) which is \(6\) Aditya Parson · 4 years, 3 months ago

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@Aditya Parson I thought the minimum of P is 6 because from the equation P=2X^2 + (3-X)^2=3X^2-6X+9 by completing the square P=3(X-1)^2 + 6, hence the minimum value is 6. Even if we consider your answer that 0 is minimum then log x base 2 has to be zero > then x=1 and repeating the same argument we find that y=1 and unfortunately this contradicts the statement that xy=8 because if minimum P is 0 then xy has to be 1. Pralhlad Hardman · 4 years, 3 months ago

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@Pralhlad Hardman You are right. Aditya Parson · 4 years, 3 months ago

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@Aditya Parson from what P=2X^2+(3−X)^2 ? Fadlan Semeion · 4 years, 3 months ago

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@Fadlan Semeion The questions asks us to express P in terms of X. Aditya Parson · 4 years, 3 months ago

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@Aditya Parson i understood about 2x^2, but i have question, from what (3−X)^2 ? Fadlan Semeion · 4 years, 3 months ago

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@Fadlan Semeion it is given xy=8 so you can derive y in terms of X as well. Aditya Parson · 4 years, 3 months ago

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