Find the least value of the function $y = 7 + 3^{x} + {2.3}^{3 - x} - x ln 27 - 9.$

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Solution

$y = 7 + 3^{x} + 2 \times 3^{3 - x} - x l n 27 - 9$
$y^{'} = 0 + 3^{x} l n 3 + 2 l n 3 (3^{3 - x}) (- 1) - 3 l n 3 - 0$
$= 3^{x} l n 3 - \frac{2 l n 3 (3^{3})}{3^{x}} - 3 l n 3 = 0$
$= (3^{x})^{2} ln 3 - 2\times 27 \times ln 3 - {3^{x}}3ln 3 = 0$
$\Rightarrow (3^{x})^{2} - 2 \times 27 - 3 \times 3^{x} = 0$
$\Rightarrow (3^{x})^{2} - 3 \times 3^{x} - 54 = 0$
$\Rightarrow (3^{x})^{2} - 9 \times 3^{x} + 6 \times 3^{x} - 54 = 0$ [Solving Quadratic Equation)
$\Rightarrow 3^{x} [3^{x} - 9] + 6 [3^{x} - 9] = 0$
$\Rightarrow (3^{x} + 6) (3^{x} - 9) = 0$
$3^{x} = - 6$ (Not possible) and,
$3^{x} - 9 = 0 \Rightarrow x = 2$
$y^{''} = (l n 3)^{2} 3^{x} + 2 (l n 3)^{2} 3^{3 - x} - 0$
Put $x = 2$
$= (l n 3)^{2} 9 + 6 (l n 3)^{2}$
$= 15 (l n 3)^{2} > 0$
Therefore, $x = 2$ is the point of minima.
at $x = 2, y = 13 - 6 l n 3$ .

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