Question

alguém poderia me ajudar!!

Answer 1

Explicação passo-a-passo:

Oi, Td bem?

A)

$2 {}^{x + 7} = \dfrac{1}{64}$

$2 {}^{x + 7} = 2 {}^{ - 6}$

$x + 7 = - 6$

$x = - 6 - 7$

x=>-13

B)

$25 {}^{x} = \sqrt{5}$

$5 {}^{2x} = 5 {}^{ \frac{1}{2} }$

$2x = \dfrac{1}{2}$

$2x \div 2 = \dfrac{1}{2} \times \dfrac{1}{2}$

x=>1/4

C)

$4 {}^{x} = 16$

$4 {}^{x} = 4 {}^{2}$

x=>2

D)

$( \dfrac{1}{9} ) {}^{x} = \sqrt{27}$

$3 {}^{ - 2x} = 3 {}^{ \frac{3}{2} }$

$- 2x = \dfrac{3}{2}$

$- 2x \div 2 = \dfrac{3}{2} \times \dfrac{1}{2}$

$- x = \dfrac{3}{4} ( - 1)$

x=>-3/4

E)

$49 {}^{2x + 3} = 343$

$7 {}^{4x + 6} = 7 {}^{3}$

$4x + 6 = 3$

$4x = 3 - 6$

$4x = - 3$

x=>-3/4

F)

$9 {}^{x} = 1$

$9 {}^{x} = 9 {}^{0}$

x=>0

G)

$5 {}^{x} = 125$

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

x=>3

H)

$5 {}^{x - 1} + 5 {}^{x + 1} = 26$

$(1 + 5 {}^{2} ) \times 5 {}^{x - 1} = 26$

$(1 + 25) \times 5 {}^{x - 1} = 26$

$26 \times 5 {}^{x - 1} = 26$

$5 {}^{x - 1} = 1$

$5 {}^{x - 1} = 5 {}^{0}$

$x - 1 = 0$

$x = 0 + 1$

x=>1

I)

$7 {}^{5x - 9} = 7 {}^{ - x + 1}$

$5x - 9 = - x + 1$

$5x + x = 1 + 9$

$6x = 10$

$x = \dfrac{10 {}^{ \div 2} }{6 {}^{ \div 2} }$

x=>5/3

J)

$( \dfrac{1}{4} ) {}^{x} = \dfrac{1}{16}$

$2 {}^{ - 2x} = 2 {}^{ - 4}$

$- 2x = - 4$

$x = \dfrac{ - 4}{ - 2}$

x=>2

K)

$3 {}^{x} = 27$

$3 {}^{x} = 3 {}^{3}$

x=>3

L)

$2 {}^{x} = \dfrac{1}{32}$

$2 {}^{x} = 2 {}^{ - 5}$

x=>-5