Question

Resolva as equações abaixo:

A)
${2}^{x} = 64$

B)
$27 = {3}^{x}$

C)
$81 = {3}^{3}$

D)
$625 = {5}^{x}$

E)
${5}^{x} = \frac{1}{125}$

F)
${3}^{x} = \frac{1}{9}$

G)
$( \frac{1}{27} ) {}^{x} = \frac{1}{9}$

H)
${12}^{x} = 1$

I)
${27}^{2x} = 9$

J)
${5}^{2x - 1} = \frac{1}{5}$

K)
$( \frac{2}{3}) {}^{3x + 1} = 1$

L)
$\frac{1}{3 {}^{x} } = \sqrt[4]{27}$
​

Answer 1

a)

${2}^{x} = 64 \\ {2}^{x} = {2}^{6} \\ x = 6$

b)

$27 = {3}^{x} \\ {3}^{x} = {3}^{3} \\ x = 3$

c) não é equação

d)

$625 = {5}^{x} \\ {5}^{x} = 625 \\ {5}^{x} = {5}^{4} \\ x = 4$

e)

${5}^{x} = \frac{1}{125} \\ {5}^{x} = {5}^{ - 3} \\ x = - 3$

f)

${3}^{x} = \frac{1}{9} \\ {3}^{x} = {3}^{ - 2} \\ x = - 2$

g)

${( \frac{1}{27} )}^{x} = \frac{1}{9} \\ {3}^{ - 3x} = {3}^{ - 2} \\ - 3x = - 2 \times ( - 1) \\ 3x = 2$

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

h)

${12}^{x} = 1 \\ {12}^{x} = {12}^{0} \\ x = 0$

i)

${27}^{2x} = 9 \\ {3}^{6x} = {3}^{2} \\ 6x = 2 \\ x = \frac{2 \div 2}{6 \div 2}$

$x = \frac{1}{ 3}$

j)

${5}^{2x - 1} = \frac{1}{5} \\ {5}^{2x - 1} = {5}^{ - 1} \\ 2x - 1 = - 1 \\ 2x = 0 \\ x = 0$

k)

${( \frac{2}{3} )}^{3x + 1} = 1 \\ {( \frac{2}{3} )}^{3x + 1} = {( \frac{2}{3} )}^{0}$

$3x + 1 = 0 \\ 3x = - 1 \\ x = - \frac{1}{3}$

l)

1/3^x=27^(¼)

3^(-x) =3^(3/4)

-x=3/4×(-1)

x=-3/4