Solve a) e^(2x) −e^(x+1) −e^x +e<0 b)4.2^(2x) −9.2^x <−2 c)9^x −4.3^(x+1) +27>0


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Question Number 70066 by Maclaurin Stickker last updated on 30/Sep/19

$S o l v e$ $a) e^{2 x} - e^{x + 1} - e^{x} + e < 0$ $b) 4 . 2^{2 x} - 9 . 2^{x} < - 2$ $c) 9^{x} - 4 . 3^{x + 1} + 27 > 0$
	Answered by Rio Michael last updated on 30/Sep/19
	$b) 4(2^x )^2 −9(2^x ) + 2 < 0 4(2^x )^2 −8(2^x ) −(2^x ) + 2 <0 4(2^x )(2^x −2) −1(2^x −2)<0 (2^x −2)[4(2^x )−1] <0 the zeros are 2^x = 2 or 2^x = (1/4) ⇒ x = 1 or x = −2 2^x <(1/4) (1/4)<2^x < 2 2^x >2 ++++ −−−−− ++++ solution set S = {2^x : (1/4)<2^x <2} or S = {x: −2 < x < 1}$
	$b) 4 {(2^{x})}^{2} - 9 (2^{x}) + 2 < 0$ $4 {(2^{x})}^{2} - 8 (2^{x}) - (2^{x}) + 2 < 0$ $4 (2^{x}) (2^{x} - 2) - 1 (2^{x} - 2) < 0$ $(2^{x} - 2) [4 (2^{x}) - 1] < 0$ $t h e z e r o s a r e 2^{x} = 2 o r 2^{x} = \frac{1}{4}$ $\Rightarrow x = 1 o r x = - 2$ $2^{x} < \frac{1}{4} \frac{1}{4} < 2^{x} < 2 2^{x} > 2$ $+ + + + - - - - - + + + +$ $s o l u t i o n s e t S = {2^{x} : \frac{1}{4} < 2^{x} < 2} o r S = {x : - 2 < x < 1}$
	Commented byMaclaurin Stickker last updated on 01/Oct/19

	$G r e a t!$
	Commented byRio Michael last updated on 02/Oct/19

	$t h a n k s$
	Answered by MJS last updated on 30/Sep/19

	$a)$ $e^{2 x} - (e + 1) e^{x} + e = 0$ $\Rightarrow e^{x} = 1 \lor e^{x} = e \Rightarrow x = 0 \lor x = 1$ $\Rightarrow 0 < x < 1$ $b)$ $2^{2 x} - \frac{9}{4} 2^{x} + \frac{1}{2} = 0$ $\Rightarrow 2^{x} = \frac{1}{4} \lor 2^{x} = 2 \Rightarrow x = - 2 \lor x = 1$ $\Rightarrow - 2 < x < 1$ $c)$ $3^{2 x} - 12 \times 3^{x} + 27 = 0$ $\Rightarrow 3^{x} = 3 \lor 3^{x} = 9 \Rightarrow x = 1 \lor x = 2$ $\Rightarrow x < 1 \lor x > 2$
	Commented byMr. K last updated on 01/Oct/19

	$A m a z i n g$
	Commented byMaclaurin Stickker last updated on 01/Oct/19

	$G r e a t!$
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