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Question Number 79127 by ~blr237~ last updated on 22/Jan/20

Solve on R∗R the following system {_(9^A +9^B +9^C =1) ^(3^A +3^B +3^C =(√3))

$$\:{Solve}\:\:{on}\:\mathbb{R}\ast\mathbb{R}\:\:{the}\:{following}\:{system} \\ $$$$\left\{_{\mathrm{9}^{{A}} +\mathrm{9}^{{B}} +\mathrm{9}^{{C}} =\mathrm{1}} ^{\mathrm{3}^{{A}} +\mathrm{3}^{{B}} +\mathrm{3}^{{C}} =\sqrt{\mathrm{3}}} \:\:\:\right. \\ $$

Commented by jagoll last updated on 23/Jan/20

because x>0 , y>0 and z>0

$$\mathrm{because}\:\mathrm{x}>\mathrm{0}\:,\:\mathrm{y}>\mathrm{0}\:\mathrm{and}\:\mathrm{z}>\mathrm{0} \\ $$

Commented by jagoll last updated on 23/Jan/20

x+y+z=(√3)(xy+xz+yz) x(y(√3)−1)+y(z(√3)−1)+z(x(√3)−1)=0 ⇒y(√3)−1=0 z(√3)−1=0 x(√3)−1=0

$$\mathrm{x}+\mathrm{y}+\mathrm{z}=\sqrt{\mathrm{3}}\left(\mathrm{xy}+\mathrm{xz}+\mathrm{yz}\right) \\ $$$$\mathrm{x}\left(\mathrm{y}\sqrt{\mathrm{3}}−\mathrm{1}\right)+\mathrm{y}\left(\mathrm{z}\sqrt{\mathrm{3}}−\mathrm{1}\right)+\mathrm{z}\left(\mathrm{x}\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{y}\sqrt{\mathrm{3}}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{z}\sqrt{\mathrm{3}}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}\sqrt{\mathrm{3}}−\mathrm{1}=\mathrm{0} \\ $$

Commented by jagoll last updated on 23/Jan/20

9^a +9^b +9^c =(3^a +3^b +3^c )^2 −2(3^(a+b) +3^(a+c) +3^(b+c) ) 1=3−2(3^(a+b) +3^(a+c) +3^(b+c) ) 3^(a+b) +3^(a+c) +3^(b+c) =1

$$\mathrm{9}^{\mathrm{a}} +\mathrm{9}^{\mathrm{b}} +\mathrm{9}^{\mathrm{c}} =\left(\mathrm{3}^{\mathrm{a}} +\mathrm{3}^{\mathrm{b}} +\mathrm{3}^{\mathrm{c}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}^{\mathrm{a}+\mathrm{b}} +\mathrm{3}^{\mathrm{a}+\mathrm{c}} +\mathrm{3}^{\mathrm{b}+\mathrm{c}} \right) \\ $$$$\mathrm{1}=\mathrm{3}−\mathrm{2}\left(\mathrm{3}^{\mathrm{a}+\mathrm{b}} +\mathrm{3}^{\mathrm{a}+\mathrm{c}} +\mathrm{3}^{\mathrm{b}+\mathrm{c}} \right) \\ $$$$\mathrm{3}^{\mathrm{a}+\mathrm{b}} +\mathrm{3}^{\mathrm{a}+\mathrm{c}} +\mathrm{3}^{\mathrm{b}+\mathrm{c}} =\mathrm{1} \\ $$

Commented by jagoll last updated on 23/Jan/20

let 3^a =x , 3^b =y , 3^c =z xy+xz+yz=1 x+y+z=(√3) x^2 +y^2 +z^2 =1

$$\mathrm{let}\:\mathrm{3}^{\mathrm{a}} =\mathrm{x}\:,\:\mathrm{3}^{\mathrm{b}} =\mathrm{y}\:,\:\mathrm{3}^{\mathrm{c}} =\mathrm{z} \\ $$$$\mathrm{xy}+\mathrm{xz}+\mathrm{yz}=\mathrm{1} \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}=\sqrt{\mathrm{3}} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{1} \\ $$

Commented by jagoll last updated on 23/Jan/20

x^2 −xy+y^2 −yz+z^2 −xz=0 x(x−y)+x(y−z)+z(z−x)=0

$$\mathrm{x}^{\mathrm{2}} −\mathrm{xy}+\mathrm{y}^{\mathrm{2}} −\mathrm{yz}+\mathrm{z}^{\mathrm{2}} −\mathrm{xz}=\mathrm{0} \\ $$$$\mathrm{x}\left(\mathrm{x}−\mathrm{y}\right)+\mathrm{x}\left(\mathrm{y}−\mathrm{z}\right)+\mathrm{z}\left(\mathrm{z}−\mathrm{x}\right)=\mathrm{0} \\ $$

Commented by jagoll last updated on 23/Jan/20

x=y=z x^2 +x^2 +x^2 =1 ⇒x=(1/(√3)) ∴ x=y=z=(1/(√3))

$$\mathrm{x}=\mathrm{y}=\mathrm{z} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$$\therefore\:\mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$

Commented by ~blr237~ last updated on 23/Jan/20

your explanation of x=y=z is not clear sir cause you don′t know if x(y−z)>0 or same for the rest

$${your}\:{explanation}\:{of}\:\:{x}={y}={z}\:{is}\:{not}\:{clear}\:{sir}\: \\ $$$${cause}\:{you}\:{don}'{t}\:{know}\:{if}\:\:{x}\left({y}−{z}\right)>\mathrm{0}\:{or}\:{same}\:{for}\:{the}\:{rest} \\ $$

Commented by jagoll last updated on 23/Jan/20

impossible x(y−z)>0 sir. because sum the three terms is zero

$$\mathrm{impossible}\:\mathrm{x}\left(\mathrm{y}−\mathrm{z}\right)>\mathrm{0}\:\mathrm{sir}. \\ $$$$\mathrm{because}\:\mathrm{sum}\:\mathrm{the}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{is}\: \\ $$$$\mathrm{zero} \\ $$

Commented by ~blr237~ last updated on 23/Jan/20

you should prove that the three of them are positive before concluding

$${you}\:{should}\:{prove}\:{that}\:{the}\:{three}\:{of}\:{them}\:{are}\:{positive}\:{before}\:{concluding} \\ $$

Commented by MJS last updated on 23/Jan/20

it must be R×R×R not R×R

$$\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathbb{R}×\mathbb{R}×\mathbb{R}\:\mathrm{not}\:\mathbb{R}×\mathbb{R} \\ $$

Answered by ~blr237~ last updated on 23/Jan/20

let find A=(3^A −(1/(√3)))^2 +(3^B −(1/(√3)))^2 +(3^C −(1/(√3)))^2 A=(9^A −(2/(√3))3^A +(1/3))+(9^B −(2/(√3))3^B +(1/3))+(9^C −(2/(√3))3^C +(1/3)) = 9^A +9^B +9^C −(2/(√3))(3^A +3^B +3^C )+1 =1−2+1=0 such as each term of A is positive we got 3^A =3^B =3^C =(1/(√3)) Finaly A=B=C=−(1/2)

$${let}\:\:{find}\:{A}=\left(\mathrm{3}^{{A}} −\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\mathrm{3}^{{B}} −\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\mathrm{3}^{{C}} −\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \\ $$$${A}=\left(\mathrm{9}^{{A}} −\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\mathrm{3}^{{A}} +\frac{\mathrm{1}}{\mathrm{3}}\right)+\left(\mathrm{9}^{{B}} −\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\mathrm{3}^{{B}} +\frac{\mathrm{1}}{\mathrm{3}}\right)+\left(\mathrm{9}^{{C}} −\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\mathrm{3}^{{C}} +\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:=\:\mathrm{9}^{{A}} +\mathrm{9}^{{B}} +\mathrm{9}^{{C}} −\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left(\mathrm{3}^{{A}} +\mathrm{3}^{{B}} +\mathrm{3}^{{C}} \right)+\mathrm{1} \\ $$$$\:\:\:=\mathrm{1}−\mathrm{2}+\mathrm{1}=\mathrm{0} \\ $$$${such}\:{as}\:{each}\:{term}\:{of}\:{A}\:{is}\:{positive}\:{we}\:{got}\: \\ $$$$\mathrm{3}^{{A}} =\mathrm{3}^{{B}} =\mathrm{3}^{{C}} =\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\: \\ $$$${Finaly}\:\:{A}={B}={C}=−\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$

Commented by jagoll last updated on 23/Jan/20

what wrong my answer?

$$\mathrm{what}\:\mathrm{wrong}\:\mathrm{my}\:\mathrm{answer}? \\ $$

Commented by ~blr237~ last updated on 23/Jan/20

we don′t know if y(√3) −1>0 , even z(√3)−1 and y(√3) −1 the proposition you want to is when U≥0,V≥0 U+V=0 ⇒ U=0 and V=0 but you did verify if firstly U,V ( there it′s x(y(√3) −1) ,y(z(√3) −1),z(x(√3) −1) ) were positive

$${we}\:{don}'{t}\:{know}\:{if}\:\:{y}\sqrt{\mathrm{3}}\:−\mathrm{1}>\mathrm{0}\:,\:{even}\:\:{z}\sqrt{\mathrm{3}}−\mathrm{1}\:{and}\:{y}\sqrt{\mathrm{3}}\:−\mathrm{1} \\ $$$${the}\:{proposition}\:{you}\:{want}\:{to}\:{is}\: \\ $$$${when}\:{U}\geqslant\mathrm{0},{V}\geqslant\mathrm{0} \\ $$$${U}+{V}=\mathrm{0}\:\Rightarrow\:{U}=\mathrm{0}\:{and}\:{V}=\mathrm{0} \\ $$$${but}\:{you}\:{did}\:{verify}\:{if}\:{firstly}\:\:{U},{V}\:\left(\:{there}\:{it}'{s}\:\:{x}\left({y}\sqrt{\mathrm{3}}\:−\mathrm{1}\right)\:,{y}\left({z}\sqrt{\mathrm{3}}\:−\mathrm{1}\right),{z}\left({x}\sqrt{\mathrm{3}}\:−\mathrm{1}\right)\:\right)\:{were}\:{positive}\: \\ $$

Commented by MJS last updated on 23/Jan/20

A=(3^A −(1/(√3)))^2 +(3^B −(1/(√3)))^2 +(3^C −(1/(√3)))^2 don′t do that, it′s confusing and simply wrong

$${A}=\left(\mathrm{3}^{{A}} −\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\mathrm{3}^{{B}} −\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\mathrm{3}^{{C}} −\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{do}\:\mathrm{that},\:\mathrm{it}'\mathrm{s}\:\mathrm{confusing}\:\mathrm{and}\:\mathrm{simply}\:\mathrm{wrong} \\ $$

Answered by MJS last updated on 23/Jan/20

3^A =x, 3^B =y, 3^C =z; x>0∧y>0∧z>0 { ((x+y+z=(√3))),((x^2 +y^2 +z^2 =1)) :} { ((z=(√3)−(x+y))),((x^2 +xy+y^2 −(√3)x−(√3)y+1=0)) :} let y=sx; x>0 { ((z=(√3)−(s+1)x)),((s^2 +((x−(√3))/x)s+((x^2 −(√3)x+1)/x^2 )=0)) :} let s=t−((x−(√3))/(2x)) { ((z=((√3)/2)−(t+(1/2))x)),((t^2 =−(((3x−(√3))^2 )/(12x^2 )))) :} ⇒ t=±(((√3)x−1)/(2x))i ⇒ s=−((x−(√3))/(2x))±(((√3)x−1)/(2x))i ⇒ y=−((x−(√3))/2)±(((√3)x−1)/2)i ⇒ z=−((x−(√3))/2)∓(((√3)x−1)/2)i ⇒ we only get a real solution if (((√3)x−1)/2)=0 ⇒ x=((√3)/3) ⇒ s=1∧t=0∧x=y=z=((√3)/3) ⇒ A=B=C=−(1/2)

$$\mathrm{3}^{{A}} ={x},\:\mathrm{3}^{{B}} ={y},\:\mathrm{3}^{{C}} ={z};\:{x}>\mathrm{0}\wedge{y}>\mathrm{0}\wedge{z}>\mathrm{0} \\ $$$$\begin{cases}{{x}+{y}+{z}=\sqrt{\mathrm{3}}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$$\begin{cases}{{z}=\sqrt{\mathrm{3}}−\left({x}+{y}\right)}\\{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}−\sqrt{\mathrm{3}}{y}+\mathrm{1}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{let}\:{y}={sx};\:{x}>\mathrm{0} \\ $$$$\begin{cases}{{z}=\sqrt{\mathrm{3}}−\left({s}+\mathrm{1}\right){x}}\\{{s}^{\mathrm{2}} +\frac{{x}−\sqrt{\mathrm{3}}}{{x}}{s}+\frac{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{0}}\end{cases} \\ $$$$\mathrm{let}\:{s}={t}−\frac{{x}−\sqrt{\mathrm{3}}}{\mathrm{2}{x}} \\ $$$$\begin{cases}{{z}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right){x}}\\{{t}^{\mathrm{2}} =−\frac{\left(\mathrm{3}{x}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{12}{x}^{\mathrm{2}} }}\end{cases} \\ $$$$\Rightarrow\:{t}=\pm\frac{\sqrt{\mathrm{3}}{x}−\mathrm{1}}{\mathrm{2}{x}}\mathrm{i} \\ $$$$\Rightarrow\:{s}=−\frac{{x}−\sqrt{\mathrm{3}}}{\mathrm{2}{x}}\pm\frac{\sqrt{\mathrm{3}}{x}−\mathrm{1}}{\mathrm{2}{x}}\mathrm{i} \\ $$$$\Rightarrow\:{y}=−\frac{{x}−\sqrt{\mathrm{3}}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}{x}−\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow\:{z}=−\frac{{x}−\sqrt{\mathrm{3}}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}{x}−\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{only}\:\mathrm{get}\:\mathrm{a}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{if} \\ $$$$\frac{\sqrt{\mathrm{3}}{x}−\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\:{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow\:{s}=\mathrm{1}\wedge{t}=\mathrm{0}\wedge{x}={y}={z}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow\:{A}={B}={C}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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