If 2^x =4^y =8^z and xyz=288, then find (1/(2x))+(1/(4y))+(1/(8z))


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Question Number 113876 by Aina Samuel Temidayo last updated on 16/Sep/20

$If 2^{x} = 4^{y} = 8^{z} and xyz = 288, then find$ $\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z}$
	Answered by Olaf last updated on 16/Sep/20

	$2^{x} = 4^{y} = 8^{z}$ $\log_{2} 2^{x} = \log_{2} 4^{y} = \log_{2} 8^{z}$ $x \log_{2} 2 = y \log_{2} 2^{2} = z \log_{2} 2^{3}$ $x = 2 y = 3 z$ $x y z = 288$ $x (\frac{x}{2}) (\frac{x}{3}) = 288$ $x^{3} = 6 \times 288 = 1728$ $x = \overset{3}{} \sqrt{1728} = \overset{3}{} \sqrt{12^{3}} = 12$ $y = \frac{x}{2} = 6$ $z = \frac{x}{3} = 4$ $\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{1}{2.12} + \frac{1}{4.6} + \frac{1}{8.4}$ $= \frac{1}{24} + \frac{1}{24} + \frac{1}{32} = \frac{11}{96}$
	Commented by Aina Samuel Temidayo last updated on 16/Sep/20

	$Thanks .$
	Answered by MJS_new last updated on 16/Sep/20

	$2^{x} = {(2^{2})}^{y} = {(2^{3})}^{z} \Rightarrow y = \frac{x}{2} \land z = \frac{x}{3}$ $x y z = \frac{x^{3}}{6} = 288 \Rightarrow x = 12$ $\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{11}{8 x} = \frac{11}{96}$
	Answered by john santu last updated on 16/Sep/20
	$(1)2^x =2^(2y) ⇒x=2y (2)2^x =2^(3z) ⇒x=3z (3)2^(2y) =2^(3z) ⇒y=((3z)/2) (4)xyz = 288⇒x((x/2))((x/3))=12^2 .2 ⇒ x^3 =12^3 → { ((x=12)),((y=6 )),((z=4)) :} therefore (1/(2x))+(1/(4y))+(1/(8z))=(1/(24))+(1/(24))+(1/(32)) = (2/(24))+(1/(32))=(2/(8.3))+(1/(8.4)) = ((8+3)/(8.3.4)) = ((11)/(96))$
	$(1) 2^{x} = 2^{2 y} \Rightarrow x = 2 y$ $(2) 2^{x} = 2^{3 z} \Rightarrow x = 3 z$ $(3) 2^{2 y} = 2^{3 z} \Rightarrow y = \frac{3 z}{2}$ $(4) x y z = 288 \Rightarrow x (\frac{x}{2}) (\frac{x}{3}) = 12^{2} . 2$ $\Rightarrow x^{3} = 12^{3} \to {\begin{cases} x = 12 \\ y = 6 \\ z = 4 \end{cases}$ $t h e r e f o r e \frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{1}{24} + \frac{1}{24} + \frac{1}{32}$ $= \frac{2}{24} + \frac{1}{32} = \frac{2}{8.3} + \frac{1}{8.4} = \frac{8 + 3}{8.3.4} = \frac{11}{96}$
	Commented by Aina Samuel Temidayo last updated on 16/Sep/20

	$Thanks .$
	Answered by floor(10²Eta[1]) last updated on 16/Sep/20

	$x = 2 y = 3 z$ $xyz = 3 z . \frac{3 z}{2} . z = \frac{{9 z}^{3}}{2} = 288 \Rightarrow z = 4$ $\Rightarrow x = 12 and y = 6$ $\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{1}{24} + \frac{1}{24} + \frac{1}{32} = \frac{11}{96}$
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