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Question Number 201162 by Calculusboy last updated on 01/Dec/23

	Answered by Frix last updated on 01/Dec/23

	$x = y = z = 1$
	Answered by Rasheed.Sindhi last updated on 01/Dec/23
	${ ((3^x +3^y +3^z =9.....(i))),((9^x +9^y +9^z =27....(ii))),((x^z +z^y +y^x =3.....(iii))) :} (i)^2 : (3^x +3^y +3^z )^2 =9^2 9^x +9^y +9^z +2.3^(x+y) +2.3^(y+z) +2.3^(z+x) =81 27+2(3^(x+y) +3^(y+z) +3^(z+x) )=81 2(3^(x+y) +3^(y+z) +3^(z+x) )=54 3^(x+y) +3^(y+z) +3^(z+x) =27 9^((x+y)/2) +9^((y+z)/2) +9^((z+x)/2) =27 Comparing with (ii) ((x+y)/2)=x∧((y+z)/2)=y∧((z+x)/2)=z −x+y=0 ∧ −y+z=0 ∧ −z+x=0 x=y=z (iii)⇒x^x +x^x +x^x =3 3x^x =3⇒x^x =1⇒x=1 x=y=z=1✓$
	${\begin{cases} 3^{x} + 3^{y} + 3^{z} = 9 . . . . . (i) \\ 9^{x} + 9^{y} + 9^{z} = 27 . . . . (i i) \\ x^{z} + z^{y} + y^{x} = 3 . . . . . (i i i) \end{cases}$ ${(i)}^{2} : {(3^{x} + 3^{y} + 3^{z})}^{2} = 9^{2}$ $9^{x} + 9^{y} + 9^{z} + 2 . 3^{x + y} + 2 . 3^{y + z} + 2 . 3^{z + x} = 81$ $27 + 2 (3^{x + y} + 3^{y + z} + 3^{z + x}) = 81$ $2 (3^{x + y} + 3^{y + z} + 3^{z + x}) = 54$ $3^{x + y} + 3^{y + z} + 3^{z + x} = 27$ $9^{\frac{x + y}{2}} + 9^{\frac{y + z}{2}} + 9^{\frac{z + x}{2}} = 27$ $C o m p a r i n g w i t h (i i)$ $\frac{x + y}{2} = x \land \frac{y + z}{2} = y \land \frac{z + x}{2} = z$ $- x + y = 0 \land - y + z = 0 \land - z + x = 0$ $x = y = z$ $(i i i) \Rightarrow x^{x} + x^{x} + x^{x} = 3$ $3 x^{x} = 3 \Rightarrow x^{x} = 1 \Rightarrow x = 1$ $x = y = z = 1 ✓$
	Commented by Calculusboy last updated on 01/Dec/23

	$t h a n k s s i r$
	Answered by witcher3 last updated on 01/Dec/23

	$a^{2} + b^{2} + c^{2} ⩾ ab + bc + ac cauchy shwartz$ $\Rightarrow 3 (a^{2} + b^{2} + c^{2}) ⩾ {(a + b + c)}^{2} equality if a = b = c$ $\Rightarrow 3.27 = 3 (3^{2 x} + 3^{2 y} + 3^{2 z}) ⩾ {(3^{x} + 3^{y} + 3^{z})}^{2} = 81$ $\Rightarrow x = y = z$ $3 rd equation \Rightarrow {3 x}^{x} = 3 \Rightarrow \forall x \in R (x^{x} = 1 \Leftrightarrow x = 1)$ $x = y = z = 1$
	Commented by Calculusboy last updated on 01/Dec/23

	$t h a n k d s i r$
	Answered by Rasheed.Sindhi last updated on 01/Dec/23
	${ ((3^x +3^y +3^z =9)),((9^x +9^y +9^z =27)),((x^z +z^y +y^x =3)) :} { ((3^(x−1) +3^(y−1) +3^(z−1) =3)),((9^(x−1) +9^(y−1) +9^(z−1) =3)),((x^z +z^y +y^x =3)) :} { ((3^(x−1) +3^(y−1) +3^(z−1) =3.....(i))),((3^(2(x−1)) +3^(2(y−1)) +3^(2(z−1)) =3...(ii))),((x^z +z^y +y^x =3.....(iii))) :} Comparing (i) & (ii) 2(x−1)=x−1⇒x−1=0⇒x=1 2(y−1)=y−1⇒y−1=0⇒y=1 2(z−1)=z−1⇒y−1=0⇒z=1$
	${\begin{cases} 3^{x} + 3^{y} + 3^{z} = 9 \\ 9^{x} + 9^{y} + 9^{z} = 27 \\ x^{z} + z^{y} + y^{x} = 3 \end{cases}$ ${\begin{cases} 3^{x - 1} + 3^{y - 1} + 3^{z - 1} = 3 \\ 9^{x - 1} + 9^{y - 1} + 9^{z - 1} = 3 \\ x^{z} + z^{y} + y^{x} = 3 \end{cases}$ ${\begin{cases} 3^{x - 1} + 3^{y - 1} + 3^{z - 1} = 3 . . . . . (i) \\ 3^{2 (x - 1)} + 3^{2 (y - 1)} + 3^{2 (z - 1)} = 3 . . . (i i) \\ x^{z} + z^{y} + y^{x} = 3 . . . . . (i i i) \end{cases}$ $C o m p a r i n g (i) & (i i)$ $2 (x - 1) = x - 1 \Rightarrow x - 1 = 0 \Rightarrow x = 1$ $2 (y - 1) = y - 1 \Rightarrow y - 1 = 0 \Rightarrow y = 1$ $2 (z - 1) = z - 1 \Rightarrow y - 1 = 0 \Rightarrow z = 1$
	Commented by Rasheed.Sindhi last updated on 01/Dec/23
	$You can compare in any way for example { ((2(x−1)=y−1⇒2x−y=1)),((2(x−1)=z−1⇒2x−z=1)) :}⇒y=z By same arguments x=y=z (iii)⇒x^x +x^x +x^x =3 3x^x =3⇒x^x =1⇒x=1 ∴ x=y=z=1$
	$Y o u c a n c o m p a r e i n a n y w a y$ $f o r e x a m p l e$ ${\begin{cases} 2 (x - 1) = y - 1 \Rightarrow 2 x - y = 1 \\ 2 (x - 1) = z - 1 \Rightarrow 2 x - z = 1 \end{cases} \Rightarrow y = z$ $B y s a m e a r g u m e n t s$ $x = y = z$ $(i i i) \Rightarrow x^{x} + x^{x} + x^{x} = 3$ $3 x^{x} = 3 \Rightarrow x^{x} = 1 \Rightarrow x = 1$ $∴ x = y = z = 1$
	Commented by Calculusboy last updated on 01/Dec/23

	$n i c e s o l u t i o n s i r$
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