Question

EQUAÇÃO EXPONENCIAL
determine o valor de x;

a) 2ˣ=64

b)7ˣx=643

c) 8ˣ-35

d) 25ˣ=625

e)9ˣ= 1/3

f) 2ˣ=1/32

g)(2/3)ˣ=8/27

h) (9/25)²ˣ=√7

i) 5ˣ=√5

j) 49ˣ=√7

k) 2ˣ⁺⁴=16

l) 5²ˣ¹=1/625

Answer 1

a)

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

b)

${7}^{x} = 643 \\ x = log_{7}(643) \\ x = \frac{ log(643) }{ log(7) } = \frac{2,80}{0,84} = \frac{10}{3}$

c)

${8}^{x} = 35 \\ x = log_{8}(35) \\ x = \frac{ log(35) }{ log(8) } = \frac{1,54}{0,90} = \frac{77}{45}$

d)

${25}^{x} = 625 \\ {( {5}^{2}) }^{x} = {5}^{4} \\ {5}^{2x} = {5}^{4} \\ 2x = 4$

$x = \frac{4}{2} = 2$

e)

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

$x = - \frac{1}{2}$

f)

${2}^{x} = \frac{1}{32} \\ {2}^{x} = {2}^{ - 5} \\ x = - 5$

g)

${( \frac{2}{3}) }^{x} = \frac{8}{27} \\ {( \frac{2}{3}) }^{x} = { (\frac{2}{3} )}^{3} \\ x = 3$

h)

${ (\frac{9}{25}) }^{2x} = \sqrt{7} \\ x = \frac{1}{2} \times log_{ \frac{9}{25} }( \sqrt{7} ) \\ x = \frac{1}{2} \times (-0,95)$

$x = -0,475$

i)

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

j)

${49}^{x} = \sqrt{7} \\ { ({7}^{2} )}^{x} = {7}^{ \frac{1}{2} } \\ {7}^{2x} = {7}^{ \frac{1}{2} } \\ 2x = \frac{1}{2}$

$4x = 1 \\ x = \frac{1}{4}$

k)

${2}^{x + 4} = 16 \\ {2}^{x + 4} = {2}^{4} \\ x + 4 = 4 \\ x = 4 - 4 \\ x = 0$

l)

${5}^{2x} = \frac{1}{625} \\ {5}^{2x} = {5}^{-4} \\ 2x = -4 \\ x = \frac{-4}{2 } = -2$