Solve-on-R-R-the-following-system-9-A-9-B-9-C-1-3-A-3-B-3-C-3-

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

S o l v e o n R * R t h e f o l l o w i n g s y s t e m

{_{9^{A} + 9^{B} + 9^{C} = 1}^{3^{A} + 3^{B} + 3^{C} = \sqrt{3}}

Commented by jagoll last updated on 23/Jan/20

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

Commented by jagoll last updated on 23/Jan/20

x + y + z = \sqrt{3} (xy + xz + yz)

x (y \sqrt{3} - 1) + y (z \sqrt{3} - 1) + z (x \sqrt{3} - 1) = 0

\Rightarrow y \sqrt{3} - 1 = 0

z \sqrt{3} - 1 = 0

x \sqrt{3} - 1 = 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

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 = \sqrt{3}

x^{2} + y^{2} + z^{2} = 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

Commented by jagoll last updated on 23/Jan/20

x = y = z

x^{2} + x^{2} + x^{2} = 1 \Rightarrow x = \frac{1}{\sqrt{3}}

∴ x = y = z = \frac{1}{\sqrt{3}}

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

y o u r e x p l a n a t i o n o f x = y = z i s n o t c l e a r s i r

c a u s e y o u {d o n}^{'} t k n o w i f x (y - z) > 0 o r s a m e f o r t h e r e s t

Commented by jagoll last updated on 23/Jan/20

impossible x (y - z) > 0 sir .

because sum the three terms is

zero

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

y o u s h o u l d p r o v e t h a t t h e t h r e e o f t h e m a r e p o s i t i v e b e f o r e c o n c l u d i n g

Commented by MJS last updated on 23/Jan/20

it must be R \times R \times R not R \times R

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

l e t f i n d A = {(3^{A} - \frac{1}{\sqrt{3}})}^{2} + {(3^{B} - \frac{1}{\sqrt{3}})}^{2} + {(3^{C} - \frac{1}{\sqrt{3}})}^{2}

A = (9^{A} - \frac{2}{\sqrt{3}} 3^{A} + \frac{1}{3}) + (9^{B} - \frac{2}{\sqrt{3}} 3^{B} + \frac{1}{3}) + (9^{C} - \frac{2}{\sqrt{3}} 3^{C} + \frac{1}{3})

= 9^{A} + 9^{B} + 9^{C} - \frac{2}{\sqrt{3}} (3^{A} + 3^{B} + 3^{C}) + 1

= 1 - 2 + 1 = 0

s u c h a s e a c h t e r m o f A i s p o s i t i v e w e g o t

3^{A} = 3^{B} = 3^{C} = \frac{1}{\sqrt{3}}

F i n a l y A = B = C = - \frac{1}{2}

Commented by jagoll last updated on 23/Jan/20

what wrong my answer ?

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

w e {d o n}^{'} t k n o w i f y \sqrt{3} - 1 > 0, e v e n z \sqrt{3} - 1 a n d y \sqrt{3} - 1

t h e p r o p o s i t i o n y o u w a n t t o i s

w h e n U ⩾ 0, V ⩾ 0

U + V = 0 \Rightarrow U = 0 a n d V = 0

b u t y o u d i d v e r i f y i f f i r s t l y U, V (t h e r e {i t}^{'} s x (y \sqrt{3} - 1), y (z \sqrt{3} - 1), z (x \sqrt{3} - 1)) w e r e p o s i t i v e

Commented by MJS last updated on 23/Jan/20

A = {(3^{A} - \frac{1}{\sqrt{3}})}^{2} + {(3^{B} - \frac{1}{\sqrt{3}})}^{2} + {(3^{C} - \frac{1}{\sqrt{3}})}^{2}

{don}^{'} t do that, {it}^{'} s confusing and simply wrong

Answered by MJS last updated on 23/Jan/20

3^{A} = x, 3^{B} = y, 3^{C} = z; x > 0 \land y > 0 \land z > 0

{\begin{cases} x + y + z = \sqrt{3} \\ x^{2} + y^{2} + z^{2} = 1 \end{cases}

{\begin{cases} z = \sqrt{3} - (x + y) \\ x^{2} + x y + y^{2} - \sqrt{3} x - \sqrt{3} y + 1 = 0 \end{cases}

let y = s x; x > 0

{\begin{cases} z = \sqrt{3} - (s + 1) x \\ s^{2} + \frac{x - \sqrt{3}}{x} s + \frac{x^{2} - \sqrt{3} x + 1}{x^{2}} = 0 \end{cases}

let s = t - \frac{x - \sqrt{3}}{2 x}

{\begin{cases} z = \frac{\sqrt{3}}{2} - (t + \frac{1}{2}) x \\ t^{2} = - \frac{{(3 x - \sqrt{3})}^{2}}{12 x^{2}} \end{cases}

\Rightarrow t = \pm \frac{\sqrt{3} x - 1}{2 x} i

\Rightarrow s = - \frac{x - \sqrt{3}}{2 x} \pm \frac{\sqrt{3} x - 1}{2 x} i

\Rightarrow y = - \frac{x - \sqrt{3}}{2} \pm \frac{\sqrt{3} x - 1}{2} i

\Rightarrow z = - \frac{x - \sqrt{3}}{2} \mp \frac{\sqrt{3} x - 1}{2} i

\Rightarrow we only get a real solution if

\frac{\sqrt{3} x - 1}{2} = 0 \Rightarrow x = \frac{\sqrt{3}}{3}

\Rightarrow s = 1 \land t = 0 \land x = y = z = \frac{\sqrt{3}}{3}

\Rightarrow A = B = C = - \frac{1}{2}

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