Find the sum of the roots of the equation: 16 ∙ 2^(∣x∣) + 5 ∙ 2^x = 42


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Question Number 146508 by mathdanisur last updated on 13/Jul/21

$F i n d t h e s u m o f t h e r o o t s o f t h e$ $e q u a t i o n :$ $16 \cdot 2^{∣ x ∣} + 5 \cdot 2^{x} = 42$
	Answered by Olaf_Thorendsen last updated on 13/Jul/21

	$16 . 2^{∣ x ∣} + 5 . 2^{x} = 42 (1)$ $1) x = 0$ $16 + 5 \neq 42 : impossible$ $2) x > 0$ $(1) : 16 . 2^{x} + 5 . 2^{x} = 42$ $2^{x} = \frac{42}{21} = 2 \Rightarrow x = 1$ $3) x < 0$ $(1) : 16 . 2^{- x} + 5 . 2^{x} = 42$ $5 . 2^{2 x} - 42 . 2^{x} + 16 = 0$ $2^{x} = \frac{42 \pm \sqrt{42^{2} - 4 \times 5 \times 16}}{10}$ $2^{x} = \frac{42 \pm 38}{10} = \frac{2}{5} or 8$ $x \ln 2 = \ln 2 - \ln 5 or 3 \ln 2$ $x = 1 - \frac{\ln 5}{\ln 2} (< 0) or 3 (> 0 : impossible)$ $Finally, x = 1 - \frac{\ln 5}{\ln 2}$ $Sum of the roots = 1 + (1 - \frac{\ln 5}{\ln 2}) = 2 - \frac{\ln 5}{\ln 2}$
	Commented by mathdanisur last updated on 13/Jul/21

	$t h a n k y o u S e r$
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