Find-the-sum-of-the-roots-of-the-equation-16-2-x-5-2-x-42-

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

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