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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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