2-to-the-power-x-2-to-the-power-x-1-3-lt-0-ease-ans-asap-

Question Number 5130 by Apoorva last updated on 16/Apr/16

2 t o t h e p o w e r x + 2 t o t h e p o w e r (- x + 1) - 3 < 0 . e a s e a n s a s a p

Commented by prakash jain last updated on 17/Apr/16

2^{{(x + 2)}^{(- x + 1) - 3}} < 0

If x \in R then 2^{{(x + 2)}^{(- x + 1) - 3}} is always > 0 .

There is no x \in R such that 2^{{(x + 2)}^{(- x + 1) - 3}} < 0

Commented by Apoorva last updated on 17/Apr/16

S o r r y s i r I m a d e t y p i n g m i s t a k e i t i s 2 t o t h e p o w e r (x + 2) + 2 t o t h e p o w e r (- x - 1) - 3 (3 i s n o t i n t h e p o w e r o f 2 i t i s s e p a r a t e) < 0 \frac{}{}

Commented by prakash jain last updated on 17/Apr/16

2^{x + 2} + 2^{(- x - 1)} - 3 < 0 ?

Pls confirm .

Commented by Apoorva last updated on 17/Apr/16

Y e s s i r y o u g o t t h e q u e s t i o n e x a c t l y c o r r e c t . P l e a s e a n s w e r i t

Answered by prakash jain last updated on 18/Apr/16

2^{x + 2} + 2^{- x - 1} - 3 < 0

2^{x} = a

4 a + \frac{1}{2 a} - 3 < 0

\frac{8 a^{2} + 1 - 6 a}{2 a} < 0

\frac{8 a^{2} - 4 a - 2 a + 1}{2 a} < 0

\frac{4 a (2 a - 1) - 1 (2 a - 1)}{2 a} < 0

\frac{(2 a - 1) (4 a - 1)}{2 a} < 0

since \frac{1}{2 a} = \frac{1}{2^{x + 1}} > 0, we need

(2 a - 1) (4 a - 1) < 0

(2 a - 1) (4 a - 1) < 0 between the two roots

a = \frac{1}{2} and a = \frac{1}{4} or

\frac{1}{4} < a < \frac{1}{2}

\frac{1}{4} < 2^{x} < \frac{1}{2}

o r - 2 < x < - 1

Commented by Apoorva last updated on 18/Apr/16

T h a n k s a l o t s i r

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