The sum of the roots of the equation x + 1 - 2 log 2 ( 3 + 2 x ) + 2 log 4 ( 10 - 2 - x ) = 0 , is

Question

The sum of the roots of the equation $x + 1 - 2 \log_{2} (3 + 2^{x}) + 2 \log_{4} (10 - 2^{- x}) = 0,$ is

Smart Achievers · Accepted Answer

(4)

$We have, x + 1 - 2 \log_{2} (3 + 2^{x}) + 2 \log_{2} (10 - 2^{- x}) = 0$

$\Rightarrow (x + 1) + \log_{2} (\frac{10 \cdot 2^{x} - 1}{2^{x}}) - \log_{2} {(3 + 2^{x})}^{2} = 0$

$\Rightarrow (x + 1) + \log_{2} [\frac{10 \cdot 2^{x} - 1}{{(3 + 2^{x})}^{2}}] - \log_{2} 2^{x} = 0$

$\Rightarrow (x + 1) + \log_{2} [\frac{10 \cdot 2^{x} - 1}{9 + {(2^{x})}^{2} + 6 \cdot 2^{x}}] - x = 0$

$\Rightarrow \log_{2} (\frac{10 \cdot 2^{x} - 1}{9 + {(2^{x})}^{2} + 6 \cdot 2^{x}}) = - 1$

$\Rightarrow (\frac{10 \cdot 2^{x} - 1}{9 + {(2^{x})}^{2} + 6 \cdot 2^{x}}) = \frac{1}{2}$

$\Rightarrow 20 \cdot 2^{x} - 2 = 9 + {(2^{x})}^{2} + 6 \cdot 2^{x} \Rightarrow {(2^{x})}^{2} - 14 \cdot 2^{x} + 11 = 0$

$\Rightarrow 2^{x} = \frac{14 \pm \sqrt{{(14)}^{2} - 4 (11) (1)}}{2} = \frac{14 \pm \sqrt{152}}{2}$

$\Rightarrow 2^{x} = 7 \pm \sqrt{38}$

$\Rightarrow x = \log_{2} (7 + \sqrt{38}) or x = \log_{2} (7 - \sqrt{38})$

$∴$ $Sum of roots = \log_{2} (7 + \sqrt{38}) + \log_{2} (7 - \sqrt{38})$

$= \log_{2} [49 - 38] = \log_{2} 11$

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