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Question Number 24576 by math solver last updated on 21/Nov/17

Commented by math solver last updated on 21/Nov/17

$sum of roots of the equation is ?$
	Answered by mrW1 last updated on 21/Nov/17

	$\frac{x + 1}{2} = \log_{2} (2^{x} + 3) - \frac{\log_{2} (1980 - 2^{- x})}{\log_{2} 4}$ $\frac{x + 1}{2} = \log_{2} (2^{x} + 3) - \frac{\log_{2} (1980 - 2^{- x})}{2}$ $x + 1 = {2 \log}_{2} (2^{x} + 3) - \log_{2} (1980 - 2^{- x})$ $x + 1 = \log_{2} \frac{{(2^{x} + 3)}^{2}}{1980 - 2^{- x}}$ $2^{x + 1} = \frac{{(2^{x} + 3)}^{2}}{1980 - 2^{- x}}$ $2 \times 2^{x} = \frac{{(2^{x} + 3)}^{2}}{1980 - 2^{- x}}$ $2 \times 1980 \times 2^{x} - 2 = {(2^{x} + 3)}^{2} = {(2^{x})}^{2} + 6 (2^{x}) + 9$ ${(2^{x})}^{2} - 3954 (2^{x}) + 11 = 0 . . . (i)$ $\Rightarrow 2^{x} = \frac{3954 \pm \sqrt{3954^{2} - 4 \times 11}}{2}$ $\Rightarrow 2^{x} = 1977 \pm \sqrt{3908518}$ $\Rightarrow x = \log_{2} (1977 \pm \sqrt{3908518})$ $(\approx - 8.49 o r 11.95)$ $\Rightarrow Σ x = \log_{2} (1977 + \sqrt{3908518}) (1977 - \sqrt{3908518})$ $\Rightarrow Σ x = \log_{2} (1977^{2} - 3908518) = \log_{2} 11 \approx 3.46$ $o r d i r e c t l y f r o m (i)$ $2^{x_{1}} \times 2^{x_{2}} = 11$ $\Rightarrow 2^{x_{1} + x_{2}} = 11$ $\Rightarrow x_{1} + x_{2} = \log_{2} 11$
	Commented by mrW1 last updated on 22/Nov/17

	$T h e r o o t s o f e q n . (i) a r e x_{1} a n d x_{2} .$ $B u t t h e e q n . (i) i s a q u a d r a t i c e q n .$ $a b o u t 2^{x}, n o t a b o u t x .$
	Commented by math solver last updated on 22/Nov/17

	$i want to ask :$ $the roots of eq . (i) are : x_{1} a n d x_{2} .$ $o r 2^{x_{1}} a n d 2^{x_{2}} ?$
	Commented by mrW1 last updated on 22/Nov/17

	$T h e o r i g i n a l q u e s t i o n a s k s h o w m u c h$ $i s x_{1} + x_{2} + . . ., n o t h o w m u c h i s$ $2^{x_{1}} + 2^{x_{2}} + . . .$
	Commented by math solver last updated on 22/Nov/17

	$thank you sir!$
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