Question

alguem me ajuda
Resolva, em IR, as equações:
a)27×2+1 = 9 5 ×
b) 534-5 = 625 ​

Answer 1

Oie, Td Bom?!

a)

$27 {}^{x {}^{2} + 1} = 9 {}^{5x}$

$(3 {}^{3} ) {}^{x {}^{2} + 1} = (3 {}^{2} ) {}^{5x}$

$3 {}^{3x {}^{2} + 3} = 3 {}^{10x}$

• Bases iguais, iguale os expoentes.

$3x {}^{2} + 3 = 10x$

$3x {}^{2} + 3 - 10x = 0$

$3x {}^{2} - 10x + 3 = 0$

$3x {}^{2} - x - 9x + 3 = 0$

$x \: . \: (3x - 1) - 3(3x - 1) = 0$

$(3x - 1) \: . \: (x - 3) = 0$

• $3x - 1 = 0⇒3x = 1⇒ x = \frac{1}{3}$

• $x - 3 = 0⇒x = 3$

$S = \left \{ \frac{1}{3} \: ,\: 3 \right \}$

b)

$5 {}^{3x - 5} = 625$

$5 {}^{3x - 5} = 5 {}^{4}$

• Bases iguais, iguale os expoentes.

$3x - 5 = 4$

$3x = 4 + 5$

$3x = 9$

$x = \frac{9}{3}$

$x = 3$

$S = \left \{ 3\right \}$

Att. Makaveli1996

Answer 2

Explicação passo-a-passo:

A)

$27 {}^{x {}^{2 + 1} } = 9 {}^{5x}$

$3 {}^{3x {}^{2} + 3 } = 3 {}^{10x}$

$3x {}^{2} + 3 = 10x$

$3x {}^{2} - 10x + 3 = 0$

$x = \dfrac{ - ( - 10) \frac{ + }{} \sqrt{( - 10) {}^{2} - 4 \times 3 \times 3 } }{2 \times 3}$

$x = \dfrac{10 \frac{ + }{} \sqrt{100 - 36} }{6}$

$x = \dfrac{10 \frac{ + }{} \sqrt{64} }{6}$

$x = \dfrac{10 \frac{ + }{}8 }{6}$

$x {}^{1} = \dfrac{10 + 8}{6} = \dfrac{18}{6} = 3$

$x {}^{2} = \dfrac{10 - 8}{6} = \dfrac{2}{6} = \dfrac{2 {}^{ \div 2} }{6 {}^{ \div 2} } = \dfrac{1}{3}$

B)

$5 {}^{3x - 5} = 625$

$5 {}^{3x - 5} = 5 {}^{4}$

$3x - 5 = 4$

$3x = 4 + 5$

$3x = 9$

$x = \dfrac{9}{3}$

$x = 3$