Solve 2^(2x) + 42^x −32=0


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Question Number 104475 by pete last updated on 21/Jul/20

$Solve 2^{2 x} + 42^{x} - 32 = 0$
	Answered by $@y@m last updated on 21/Jul/20

	$I t h i n k t h e r e i s a t y p o i n q u e s t i o n .$ $I t s h o u l d b e :$ $Solve 2^{2 x} + 4 . 2^{x} - 32 = 0$ $S o l u t i o n : L e t 2^{x} = y$ $y^{2} + 4 y - 32 = 0$ $y = 4, - 8$ $B u t y = - 8 i s i n a d m i s s i b l e .$ $∴ y = 4$ $x = 2$
	Commented by pete last updated on 21/Jul/20

	$ok thanks very much sir$
	Answered by Dwaipayan Shikari last updated on 21/Jul/20
	$2^(2x) +4.2^x −32=0 p^2 +4p−32=0 {2^x =p p=−8,4 2^x =4⇒x=2 2^x =−8 xlog2=log(−1)+log(8) x=((log(e^(πi) )+log(8))/(log(2)))=((πi+log8)/(log2))(Another solution)$
	$2^{2 x} + 4 . 2^{x} - 32 = 0$ $p^{2} + 4 p - 32 = 0 {2^{x} = p$ $p = - 8, 4$ $2^{x} = 4 \Rightarrow x = 2$ $2^{x} = - 8$ $x l o g 2 = l o g (- 1) + l o g (8)$ $x = \frac{l o g (e^{π i}) + l o g (8)}{l o g (2)} = \frac{π i + l o g 8}{l o g 2} (A n o t h e r s o l u t i o n)$
	Commented by $@y@m last updated on 22/Jul/20

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