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Question Number 17075 by tawa tawa last updated on 30/Jun/17
Given that:  log((x/(y − z))) = log((y/(z − x))) = log((z/(x − y)))  Show that :   x^x  × y^y  × z^z  = 1
$$\mathrm{Given}\:\mathrm{that}:\:\:\mathrm{log}\left(\frac{\mathrm{x}}{\mathrm{y}\:−\:\mathrm{z}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{y}}{\mathrm{z}\:−\:\mathrm{x}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{z}}{\mathrm{x}\:−\:\mathrm{y}}\right) \\ $$$$\mathrm{Show}\:\mathrm{that}\::\:\:\:\mathrm{x}^{\mathrm{x}} \:×\:\mathrm{y}^{\mathrm{y}} \:×\:\mathrm{z}^{\mathrm{z}} \:=\:\mathrm{1} \\ $$
Commented by RasheedSoomro last updated on 30/Jun/17
   log((x/(y − z))) = log((y/(z − x))) = log((z/(x − y)))   ⇒ (x/(y − z)) = (y/(z − x)) = (z/(x − y))=((x+y+z)/(y−z+z−x+x−y))                                                          =((x+y+z)/0)=∞          ???
$$\:\:\:\mathrm{log}\left(\frac{\mathrm{x}}{\mathrm{y}\:−\:\mathrm{z}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{y}}{\mathrm{z}\:−\:\mathrm{x}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{z}}{\mathrm{x}\:−\:\mathrm{y}}\right) \\ $$$$\:\Rightarrow\:\frac{\mathrm{x}}{\mathrm{y}\:−\:\mathrm{z}}\:=\:\frac{\mathrm{y}}{\mathrm{z}\:−\:\mathrm{x}}\:=\:\frac{\mathrm{z}}{\mathrm{x}\:−\:\mathrm{y}}=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{y}−\mathrm{z}+\mathrm{z}−\mathrm{x}+\mathrm{x}−\mathrm{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{0}}=\infty \\ $$$$\:\:\:\:\:\:\:\:??? \\ $$$$ \\ $$
Answered by 433 last updated on 30/Jun/17
    (x/(y−z))>0 & (y/(z−x))>0 & (z/(x−y))>0  x(y−z)>0 & y(z−x)>0 & z(x−y)>0  xy−xz>0 & yz−yx>0 & zx−zy>0  xy>xz & yz>yx & zx>zy  xy>xz>zy>yx  xy>yx
$$ \\ $$$$ \\ $$$$\frac{{x}}{{y}−{z}}>\mathrm{0}\:\&\:\frac{{y}}{{z}−{x}}>\mathrm{0}\:\&\:\frac{{z}}{{x}−{y}}>\mathrm{0} \\ $$$${x}\left({y}−{z}\right)>\mathrm{0}\:\&\:{y}\left({z}−{x}\right)>\mathrm{0}\:\&\:{z}\left({x}−{y}\right)>\mathrm{0} \\ $$$${xy}−{xz}>\mathrm{0}\:\&\:{yz}−{yx}>\mathrm{0}\:\&\:{zx}−{zy}>\mathrm{0} \\ $$$${xy}>{xz}\:\&\:{yz}>{yx}\:\&\:{zx}>{zy} \\ $$$${xy}>{xz}>{zy}>{yx} \\ $$$${xy}>{yx} \\ $$
Answered by mrW1 last updated on 30/Jun/17
This question can not be correct!    if  log((x/(y − z))) = log((y/(z − x))) = log((z/(x − y)))  ⇒(x/(y−z))=(y/(z−x))=(z/(x−y))=(1/a) let′s say  x≠0,y≠0,z≠0,a≠0  y−z=ax   ...(i)  z−x=ay   ...(ii)  x−y=az   ...(iii)    0=a(x+y+z)  ⇒x+y+z=0  y+z=−x  y−z=ax  ⇒y=((−1+a)/2)x  ⇒z=((−1−a)/2)x  put this into (ii) or (iii)  x−((−1+a)/2)x=a((−1−a)/2)x  (2+1−a)x=(−1−a)ax  (3−a+a^2 +a)x=0  a^2 +3=0  !  no such value for a is possible.  that means it′s not true that  log((x/(y − z))) = log((y/(z − x))) = log((z/(x − y)))
$$\mathrm{This}\:\mathrm{question}\:\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{correct}! \\ $$$$ \\ $$$$\mathrm{if}\:\:\mathrm{log}\left(\frac{\mathrm{x}}{\mathrm{y}\:−\:\mathrm{z}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{y}}{\mathrm{z}\:−\:\mathrm{x}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{z}}{\mathrm{x}\:−\:\mathrm{y}}\right) \\ $$$$\Rightarrow\frac{\mathrm{x}}{\mathrm{y}−\mathrm{z}}=\frac{\mathrm{y}}{\mathrm{z}−\mathrm{x}}=\frac{\mathrm{z}}{\mathrm{x}−\mathrm{y}}=\frac{\mathrm{1}}{\mathrm{a}}\:\mathrm{let}'\mathrm{s}\:\mathrm{say} \\ $$$$\mathrm{x}\neq\mathrm{0},\mathrm{y}\neq\mathrm{0},\mathrm{z}\neq\mathrm{0},\mathrm{a}\neq\mathrm{0} \\ $$$$\mathrm{y}−\mathrm{z}=\mathrm{ax}\:\:\:…\left(\mathrm{i}\right) \\ $$$$\mathrm{z}−\mathrm{x}=\mathrm{ay}\:\:\:…\left(\mathrm{ii}\right) \\ $$$$\mathrm{x}−\mathrm{y}=\mathrm{az}\:\:\:…\left(\mathrm{iii}\right) \\ $$$$ \\ $$$$\mathrm{0}=\mathrm{a}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right) \\ $$$$\Rightarrow\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{0} \\ $$$$\mathrm{y}+\mathrm{z}=−\mathrm{x} \\ $$$$\mathrm{y}−\mathrm{z}=\mathrm{ax} \\ $$$$\Rightarrow\mathrm{y}=\frac{−\mathrm{1}+\mathrm{a}}{\mathrm{2}}\mathrm{x} \\ $$$$\Rightarrow\mathrm{z}=\frac{−\mathrm{1}−\mathrm{a}}{\mathrm{2}}\mathrm{x} \\ $$$$\mathrm{put}\:\mathrm{this}\:\mathrm{into}\:\left(\mathrm{ii}\right)\:\mathrm{or}\:\left(\mathrm{iii}\right) \\ $$$$\mathrm{x}−\frac{−\mathrm{1}+\mathrm{a}}{\mathrm{2}}\mathrm{x}=\mathrm{a}\frac{−\mathrm{1}−\mathrm{a}}{\mathrm{2}}\mathrm{x} \\ $$$$\left(\mathrm{2}+\mathrm{1}−\mathrm{a}\right)\mathrm{x}=\left(−\mathrm{1}−\mathrm{a}\right)\mathrm{ax} \\ $$$$\left(\mathrm{3}−\mathrm{a}+\mathrm{a}^{\mathrm{2}} +\mathrm{a}\right)\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{3}=\mathrm{0}\:\:! \\ $$$$\mathrm{no}\:\mathrm{such}\:\mathrm{value}\:\mathrm{for}\:\mathrm{a}\:\mathrm{is}\:\mathrm{possible}. \\ $$$$\mathrm{that}\:\mathrm{means}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}\:\mathrm{that} \\ $$$$\mathrm{log}\left(\frac{\mathrm{x}}{\mathrm{y}\:−\:\mathrm{z}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{y}}{\mathrm{z}\:−\:\mathrm{x}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{z}}{\mathrm{x}\:−\:\mathrm{y}}\right) \\ $$

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