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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)) :}

$${Determine},\:{in}\:{simplest}\:{form}\:{the} \\ $$$${smallest}\:{of}\:{the}\:{three}\:{numbers}\:{x}, \\ $$$${y}\:{and}\:{z}\:{which}\:{satisfy}\:{the}\:{system} \\ $$$$\begin{cases}{\mathrm{log}\:_{\mathrm{9}} \left({x}\right)+\mathrm{log}\:_{\mathrm{9}} \left({y}\right)+\mathrm{log}\:_{\mathrm{3}} \left({z}\right)=\mathrm{2}}\\{\mathrm{log}\:_{\mathrm{16}} \left({x}\right)+\mathrm{log}\:_{\mathrm{4}} \left({y}\right)+\mathrm{log}\:_{\mathrm{16}} \left({z}\right)=\mathrm{1}}\\{\mathrm{log}\:_{\mathrm{5}} \left({x}\right)+\mathrm{log}\:_{\mathrm{25}} \left({y}\right)+\mathrm{log}\:_{\mathrm{25}} \left({z}\right)=\mathrm{0}}\end{cases} \\ $$

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))) :}

$$\rightarrow\begin{cases}{\mathrm{log}\:_{\mathrm{9}} \left({xyz}^{\mathrm{2}} \right)=\mathrm{2}\rightarrow{xyz}^{\mathrm{2}} =\mathrm{81}}\\{\mathrm{log}\:_{\mathrm{16}} \left({xy}^{\mathrm{2}} {z}\right)=\mathrm{1}\rightarrow{xy}^{\mathrm{2}} {z}=\mathrm{16}}\\{\mathrm{log}\:_{\mathrm{25}} \left({x}^{\mathrm{2}} {yz}\right)=\mathrm{0}\rightarrow{x}^{\mathrm{2}} {yz}=\mathrm{1}}\end{cases} \\ $$$$\Leftrightarrow\:\left({xyz}\right)^{\mathrm{4}} \:=\:\mathrm{81}×\mathrm{16}×\mathrm{1}=\left(\mathrm{6}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\:{xyz}\:=\:\mathrm{6}\:\begin{cases}{{z}=\frac{\mathrm{81}}{\mathrm{6}}\:=\:\frac{\mathrm{27}}{\mathrm{2}}}\\{{y}=\frac{\mathrm{16}}{\mathrm{6}}=\frac{\mathrm{8}}{\mathrm{3}}}\\{{x}=\frac{\mathrm{1}}{\mathrm{6}}}\end{cases} \\ $$

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)

$$\left(\mathrm{I}\right)\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{x}\right)}{\mathrm{2}}+\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{y}\right)}{\mathrm{2}}+\mathrm{log}_{\mathrm{3}} \left(\mathrm{z}\right)=\mathrm{2} \\ $$$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{3}} \left(\mathrm{y}\right)+\mathrm{2log}_{\mathrm{3}} \left(\mathrm{z}\right)=\mathrm{4} \\ $$$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{xyz}^{\mathrm{2}} \right)=\mathrm{4} \\ $$$$\mathrm{xyz}^{\mathrm{2}} =\mathrm{81} \\ $$$$\left(\mathrm{II}\right)\mathrm{log}_{\mathrm{4}} \left(\mathrm{x}\right)+\mathrm{2log}_{\mathrm{4}} \left(\mathrm{y}\right)+\mathrm{log}_{\mathrm{4}} \left(\mathrm{z}\right)=\mathrm{2} \\ $$$$\mathrm{log}_{\mathrm{4}} \left(\mathrm{xy}^{\mathrm{2}} \mathrm{z}\right)=\mathrm{2} \\ $$$$\mathrm{xy}^{\mathrm{2}} \mathrm{z}=\mathrm{16} \\ $$$$\left(\mathrm{III}\right)\mathrm{2log}_{\mathrm{5}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{5}} \left(\mathrm{y}\right)+\mathrm{log}_{\mathrm{5}} \left(\mathrm{z}\right)=\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} \mathrm{yz}\right)=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{yz}=\mathrm{1} \\ $$$$\begin{cases}{\mathrm{xyz}^{\mathrm{2}} =\mathrm{81}}\\{\mathrm{xy}^{\mathrm{2}} \mathrm{z}=\mathrm{16}}\\{\mathrm{x}^{\mathrm{2}} \mathrm{yz}=\mathrm{1}}\end{cases} \\ $$$$\mathrm{xyz}=\frac{\mathrm{81}}{\mathrm{z}} \\ $$$$\frac{\mathrm{81y}}{\mathrm{z}}=\mathrm{16}\Rightarrow\mathrm{16z}=\mathrm{81y}\Rightarrow\mathrm{16x}=\mathrm{y} \\ $$$$\frac{\mathrm{81x}}{\mathrm{z}}=\mathrm{1}\Rightarrow\mathrm{z}=\mathrm{81x} \\ $$$$\mathrm{x}.\mathrm{16x}.\mathrm{81x}=\frac{\mathrm{81}}{\mathrm{81x}}\Rightarrow\mathrm{x}^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{4}} }\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{y}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\mathrm{z}=\frac{\mathrm{27}}{\mathrm{2}} \\ $$

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