Question

preciso dessa resposta urgente:

a) (2^x+4)+(2^x+2)=80

b) (3^x+2)+(3^x+1) =36

Answer 1

a)
(2^(x+4)) + (2^(x+2)) = 80
(2^x · 2^4) + (2^x · 2^2) = 80
(2^x · 16) + (2^x · 4) = 80

2^x = y

(2^x · 16)+(2^x · 4) = 80
16y + 4y = 80
20y = 80
y = 80/20
y = 4

2^x = y
2^x = 4
2^x = 2^2
x = 2

b)
(3^(x + 2))+(3^(x + 1)) = 36
(3^x · 3^2)+(3^x · 3^1) =36
(3^x · 9)+(3^x · 3) = 36

3^x = y

(3^x · 9)+(3^x · 3) = 36
9y + 3y = 36
12y = 36
y = 36/12
y =3

3^x = y
3^x = 3
3^x = 3^1
x = 1

a) x = 2
b) x = 1