Question

16.)
A.
B.
C.

Algebra 2 chapter 7

Answer 1

`64^(x + 3) * 64^(3x - 2) = ((1)/(4))^(-2x + 2)`


`64^(x + 3 + 3x - 2) = (64^{-(1)/(3)})^(-2x + 2)`


`64^(4x + 1) = 64^{(2)/(3)x - (2)/(3)}`


`64^(4x + 1) = 64^{(2x - 2)/(3)}`


`4x + 1 = (2x - 2)/(3)`


`12x + 3 = 2x - 2`


`10x = -5`


`x = -(5)/(10)`


`x = -(1)/(2)`

Answer 2

#17)

64^(x+3) * 64^(3x - 2) = (1/4)^(-2x - 2)
64^(x + 3 + 3x - 2) = (4^-1)^(-2x - 2)
64^(4x+1) = 4^(2x + 2)
(4^3)^(4x+1) = 4^(2x + 2)
4^(12x + 3) = 4^(2x + 2)

So
12x + 3 = 2x + 2
10x = -1
x = -1/10
x = - 0.1

Hope it helps