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Let two positive real numbers \(x,y\) change and satisfy the relation
\(4 + \ln \frac{{2x + 2y + 1}}{{5xy}} = 20xy – \left( {8x + 8y} \right)\).
The minimum value of the expression \(P = xy + 9\) is equal to
A. \(m = 11.\) B. \(m = 10.\) C. \(m = 12 \cdot \) D. \(m = \frac{{19}}{2} \cdot \ )
The answer:
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We have
\(4 + \ln \frac{{2x + 2y + 1}}{{5xy}} = 20xy – 8x – 8y\,\,\, \Leftrightarrow \ln \left( {2x + 2y + 1} \right ) – \ln \left( {5xy} \right)\, = 4\left( {5xy} \right) – 4\left( {2x + 2y + 1} \right)\)\( \Leftrightarrow \ln \ left( {2x + 2y + 1} \right) + 4\left( {2x + 2y + 1} \right) = \ln \left( {5xy} \right)\, + 4\left( {5xy} \ right)\,\,\,\,\left( 1 \right).\)
Consider the function \(f\left( t \right) = 4t + \ln t\) with \(t > 0\).
Since \(f’\left( t \right) = 4 + \frac{1}{t} > 0,\,\,\forall t > 0\) \(f
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These are VD-VDC sentences in the topic REVIEW OF CODE FUNCTIONS – LOGARIT.
