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Question-201162

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Question Number 201162 by Calculusboy last updated on 01/Dec/23
Answered by Frix last updated on 01/Dec/23
x=y=z=1
$${x}={y}={z}=\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Dec/23
{ ((3^x +3^y +3^z =9.....(i))),((9^x +9^y +9^z =27....(ii))),((x^z +z^y +y^x =3.....(iii))) :} (i)^2 : (3^x +3^y +3^z )^2 =9^2 9^x +9^y +9^z +2.3^(x+y) +2.3^(y+z) +2.3^(z+x) =81 27+2(3^(x+y) +3^(y+z) +3^(z+x) )=81 2(3^(x+y) +3^(y+z) +3^(z+x) )=54 3^(x+y) +3^(y+z) +3^(z+x) =27 9^((x+y)/2) +9^((y+z)/2) +9^((z+x)/2) =27 Comparing with (ii) ((x+y)/2)=x∧((y+z)/2)=y∧((z+x)/2)=z −x+y=0 ∧ −y+z=0 ∧ −z+x=0 x=y=z (iii)⇒x^x +x^x +x^x =3 3x^x =3⇒x^x =1⇒x=1 x=y=z=1✓
$$\begin{cases}{\mathrm{3}^{{x}} +\mathrm{3}^{{y}} +\mathrm{3}^{{z}} =\mathrm{9}…..\left({i}\right)}\\{\mathrm{9}^{{x}} +\mathrm{9}^{{y}} +\mathrm{9}^{{z}} =\mathrm{27}….\left({ii}\right)}\\{{x}^{{z}} +{z}^{{y}} +{y}^{{x}} =\mathrm{3}…..\left({iii}\right)}\end{cases} \\ $$$$\left({i}\right)^{\mathrm{2}} :\:\left(\mathrm{3}^{{x}} +\mathrm{3}^{{y}} +\mathrm{3}^{{z}} \right)^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} \\ $$$$\mathrm{9}^{{x}} +\mathrm{9}^{{y}} +\mathrm{9}^{{z}} +\mathrm{2}.\mathrm{3}^{{x}+{y}} +\mathrm{2}.\mathrm{3}^{{y}+{z}} +\mathrm{2}.\mathrm{3}^{{z}+{x}} =\mathrm{81} \\ $$$$\mathrm{27}+\mathrm{2}\left(\mathrm{3}^{{x}+{y}} +\mathrm{3}^{{y}+{z}} +\mathrm{3}^{{z}+{x}} \right)=\mathrm{81} \\ $$$$\mathrm{2}\left(\mathrm{3}^{{x}+{y}} +\mathrm{3}^{{y}+{z}} +\mathrm{3}^{{z}+{x}} \right)=\mathrm{54} \\ $$$$\mathrm{3}^{{x}+{y}} +\mathrm{3}^{{y}+{z}} +\mathrm{3}^{{z}+{x}} =\mathrm{27} \\ $$$$\mathrm{9}^{\frac{{x}+{y}}{\mathrm{2}}} +\mathrm{9}^{\frac{{y}+{z}}{\mathrm{2}}} +\mathrm{9}^{\frac{{z}+{x}}{\mathrm{2}}} =\mathrm{27} \\ $$$${Comparing}\:{with}\:\left({ii}\right) \\ $$$$\frac{{x}+{y}}{\mathrm{2}}={x}\wedge\frac{{y}+{z}}{\mathrm{2}}={y}\wedge\frac{{z}+{x}}{\mathrm{2}}={z} \\ $$$$−{x}+{y}=\mathrm{0}\:\wedge\:−{y}+{z}=\mathrm{0}\:\wedge\:−{z}+{x}=\mathrm{0} \\ $$$${x}={y}={z} \\ $$$$\left({iii}\right)\Rightarrow{x}^{{x}} +{x}^{{x}} +{x}^{{x}} =\mathrm{3} \\ $$$$\mathrm{3}{x}^{{x}} =\mathrm{3}\Rightarrow{x}^{{x}} =\mathrm{1}\Rightarrow{x}=\mathrm{1} \\ $$$${x}={y}={z}=\mathrm{1}\checkmark \\ $$
Commented by Calculusboy last updated on 01/Dec/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by witcher3 last updated on 01/Dec/23
a^2 +b^2 +c^2 ≥ab+bc+ac cauchy shwartz ⇒3(a^2 +b^2 +c^2 )≥(a+b+c)^2 equality if a=b=c ⇒3.27=3(3^(2x) +3^(2y) +3^(2z) )≥(3^x +3^y +3^z )^2 =81 ⇒x=y=z 3rd equation⇒3x^x =3⇒∀x∈R(x^x =1⇔x=1) x=y=z=1
$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \geqslant\mathrm{ab}+\mathrm{bc}+\mathrm{ac}\:\mathrm{cauchy}\:\mathrm{shwartz} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\geqslant\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} \:\mathrm{equality}\:\mathrm{if}\:\mathrm{a}=\mathrm{b}=\mathrm{c} \\ $$$$\Rightarrow\mathrm{3}.\mathrm{27}=\mathrm{3}\left(\mathrm{3}^{\mathrm{2x}} +\mathrm{3}^{\mathrm{2y}} +\mathrm{3}^{\mathrm{2z}} \right)\geqslant\left(\mathrm{3}^{\mathrm{x}} +\mathrm{3}^{\mathrm{y}} +\mathrm{3}^{\mathrm{z}} \right)^{\mathrm{2}} =\mathrm{81} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{y}=\mathrm{z} \\ $$$$\mathrm{3rd}\:\mathrm{equation}\Rightarrow\mathrm{3x}^{\mathrm{x}} =\mathrm{3}\Rightarrow\forall\mathrm{x}\in\mathbb{R}\left(\mathrm{x}^{\mathrm{x}} =\mathrm{1}\Leftrightarrow\mathrm{x}=\mathrm{1}\right) \\ $$$$\mathrm{x}=\mathrm{y}=\mathrm{z}=\mathrm{1} \\ $$
Commented by Calculusboy last updated on 01/Dec/23
thankd sir
$$\boldsymbol{{thankd}}\:\boldsymbol{{sir}} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Dec/23
{ ((3^x +3^y +3^z =9)),((9^x +9^y +9^z =27)),((x^z +z^y +y^x =3)) :} { ((3^(x−1) +3^(y−1) +3^(z−1) =3)),((9^(x−1) +9^(y−1) +9^(z−1) =3)),((x^z +z^y +y^x =3)) :} { ((3^(x−1) +3^(y−1) +3^(z−1) =3.....(i))),((3^(2(x−1)) +3^(2(y−1)) +3^(2(z−1)) =3...(ii))),((x^z +z^y +y^x =3.....(iii))) :} Comparing (i) & (ii) 2(x−1)=x−1⇒x−1=0⇒x=1 2(y−1)=y−1⇒y−1=0⇒y=1 2(z−1)=z−1⇒y−1=0⇒z=1
$$\begin{cases}{\mathrm{3}^{{x}} +\mathrm{3}^{{y}} +\mathrm{3}^{{z}} =\mathrm{9}}\\{\mathrm{9}^{{x}} +\mathrm{9}^{{y}} +\mathrm{9}^{{z}} =\mathrm{27}}\\{{x}^{{z}} +{z}^{{y}} +{y}^{{x}} =\mathrm{3}}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{3}^{{x}−\mathrm{1}} +\mathrm{3}^{{y}−\mathrm{1}} +\mathrm{3}^{{z}−\mathrm{1}} =\mathrm{3}}\\{\mathrm{9}^{{x}−\mathrm{1}} +\mathrm{9}^{{y}−\mathrm{1}} +\mathrm{9}^{{z}−\mathrm{1}} =\mathrm{3}}\\{{x}^{{z}} +{z}^{{y}} +{y}^{{x}} =\mathrm{3}}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{3}^{{x}−\mathrm{1}} +\mathrm{3}^{{y}−\mathrm{1}} +\mathrm{3}^{{z}−\mathrm{1}} =\mathrm{3}…..\left({i}\right)}\\{\mathrm{3}^{\mathrm{2}\left({x}−\mathrm{1}\right)} +\mathrm{3}^{\mathrm{2}\left({y}−\mathrm{1}\right)} +\mathrm{3}^{\mathrm{2}\left({z}−\mathrm{1}\right)} =\mathrm{3}…\left({ii}\right)}\\{{x}^{{z}} +{z}^{{y}} +{y}^{{x}} =\mathrm{3}…..\left({iii}\right)}\end{cases}\: \\ $$$${Comparing}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\mathrm{2}\left({x}−\mathrm{1}\right)={x}−\mathrm{1}\Rightarrow{x}−\mathrm{1}=\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $$$$\mathrm{2}\left({y}−\mathrm{1}\right)={y}−\mathrm{1}\Rightarrow{y}−\mathrm{1}=\mathrm{0}\Rightarrow{y}=\mathrm{1} \\ $$$$\mathrm{2}\left({z}−\mathrm{1}\right)={z}−\mathrm{1}\Rightarrow{y}−\mathrm{1}=\mathrm{0}\Rightarrow{z}=\mathrm{1} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Dec/23
You can compare in any way for example { ((2(x−1)=y−1⇒2x−y=1)),((2(x−1)=z−1⇒2x−z=1)) :}⇒y=z By same arguments x=y=z (iii)⇒x^x +x^x +x^x =3 3x^x =3⇒x^x =1⇒x=1 ∴ x=y=z=1
$${You}\:{can}\:{compare}\:{in}\:{any}\:{way} \\ $$$${for}\:{example} \\ $$$$\begin{cases}{\mathrm{2}\left({x}−\mathrm{1}\right)={y}−\mathrm{1}\Rightarrow\mathrm{2}{x}−{y}=\mathrm{1}}\\{\mathrm{2}\left({x}−\mathrm{1}\right)={z}−\mathrm{1}\Rightarrow\mathrm{2}{x}−{z}=\mathrm{1}}\end{cases}\Rightarrow{y}={z} \\ $$$${By}\:\:{same}\:{arguments} \\ $$$${x}={y}={z} \\ $$$$\left({iii}\right)\Rightarrow{x}^{{x}} +{x}^{{x}} +{x}^{{x}} =\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}^{{x}} =\mathrm{3}\Rightarrow{x}^{{x}} =\mathrm{1}\Rightarrow{x}=\mathrm{1} \\ $$$$\therefore\:{x}={y}={z}=\mathrm{1} \\ $$
Commented by Calculusboy last updated on 01/Dec/23
nice solution sir
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{sir}} \\ $$

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