Determine, in simplest form the smallest of the three numbers x, y and z which satisfy the system { ((log _9 (x)+log _9 (y)+log _3 (z)=2)),((log _(16) (x)+log _4 (y)+log _(16) (z)=1)),((log _5 (x)+log _(25) (y)+log


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Question Number 115859 by bemath last updated on 29/Sep/20
$Determine, in simplest form the smallest of the three numbers x, y and z which satisfy the system { ((log _9 (x)+log _9 (y)+log _3 (z)=2)),((log _(16) (x)+log _4 (y)+log _(16) (z)=1)),((log _5 (x)+log _(25) (y)+log _(25) (z)=0)) :}$
$D e t e r m i n e, i n s i m p l e s t f o r m t h e$ $s m a l l e s t o f t h e t h r e e n u m b e r s x,$ $y a n d z w h i c h s a t i s f y t h e s y s t e m$ ${\begin{cases} \log_{9} (x) + \log_{9} (y) + \log_{3} (z) = 2 \\ \log_{16} (x) + \log_{4} (y) + \log_{16} (z) = 1 \\ \log_{5} (x) + \log_{25} (y) + \log_{25} (z) = 0 \end{cases}$
	Answered by bobhans last updated on 29/Sep/20
	$→ { ((log _9 (xyz^2 )=2→xyz^2 =81)),((log _(16) (xy^2 z)=1→xy^2 z=16)),((log _(25) (x^2 yz)=0→x^2 yz=1)) :} ⇔ (xyz)^4 = 81×16×1=(6)^4 ⇒ xyz = 6 { ((z=((81)/6) = ((27)/2))),((y=((16)/6)=(8/3))),((x=(1/6))) :}$
	$\to {\begin{cases} \log_{9} ({x y z}^{2}) = 2 \to {x y z}^{2} = 81 \\ \log_{16} ({x y}^{2} z) = 1 \to {x y}^{2} z = 16 \\ \log_{25} (x^{2} y z) = 0 \to x^{2} y z = 1 \end{cases}$ $\Leftrightarrow {(x y z)}^{4} = 81 \times 16 \times 1 = {(6)}^{4}$ $\Rightarrow x y z = 6 {\begin{cases} z = \frac{81}{6} = \frac{27}{2} \\ y = \frac{16}{6} = \frac{8}{3} \\ x = \frac{1}{6} \end{cases}$
	Answered by floor(10²Eta[1]) last updated on 29/Sep/20
	$(I)((log_3 (x))/2)+((log_3 (y))/2)+log_3 (z)=2 log_3 (x)+log_3 (y)+2log_3 (z)=4 log_3 (xyz^2 )=4 xyz^2 =81 (II)log_4 (x)+2log_4 (y)+log_4 (z)=2 log_4 (xy^2 z)=2 xy^2 z=16 (III)2log_5 (x)+log_5 (y)+log_5 (z)=0 log_5 (x^2 yz)=0 x^2 yz=1 { ((xyz^2 =81)),((xy^2 z=16)),((x^2 yz=1)) :} xyz=((81)/z) ((81y)/z)=16⇒16z=81y⇒16x=y ((81x)/z)=1⇒z=81x x.16x.81x=((81)/(81x))⇒x^4 =(1/6^4 )⇒x=(1/6) y=(8/3) z=((27)/2)$
	$(I) \frac{\log_{3} (x)}{2} + \frac{\log_{3} (y)}{2} + \log_{3} (z) = 2$ $\log_{3} (x) + \log_{3} (y) + {2 \log}_{3} (z) = 4$ $\log_{3} ({xyz}^{2}) = 4$ ${xyz}^{2} = 81$ $(II) \log_{4} (x) + {2 \log}_{4} (y) + \log_{4} (z) = 2$ $\log_{4} ({xy}^{2} z) = 2$ ${xy}^{2} z = 16$ $(III) {2 \log}_{5} (x) + \log_{5} (y) + \log_{5} (z) = 0$ $\log_{5} (x^{2} yz) = 0$ $x^{2} yz = 1$ ${\begin{cases} {xyz}^{2} = 81 \\ {xy}^{2} z = 16 \\ x^{2} yz = 1 \end{cases}$ $xyz = \frac{81}{z}$ $\frac{81 y}{z} = 16 \Rightarrow 16 z = 81 y \Rightarrow 16 x = y$ $\frac{81 x}{z} = 1 \Rightarrow z = 81 x$ $x . 16 x . 81 x = \frac{81}{81 x} \Rightarrow x^{4} = \frac{1}{6^{4}} \Rightarrow x = \frac{1}{6}$ $y = \frac{8}{3}$ $z = \frac{27}{2}$
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