If (1/4)^(x+y) = 256 and log₄(x-y) = 3, what are the exact values of x and y

x = 30 and y = -34

Step-by-step explanation:

Given the following functions

(1/4)^(x+y) = 256... 1

log₄(x-y) = 3 2

From equation 2;

x-y = 4³

x-y = 64

x = 64 + y ... 3

Substitutw 3 into 1

From 1:

(1/4)^(x+y) = 256

(1/4)^(64+y+y) = 256

(1/4)^(64+2y) = 256

Take log₄ of both sides

64+2y log₄ (1/4) = log₄256

-(64+2y) = 4log₄4

-(64+2y) = 4

64+2y = -4

2y = -4 - 64

2y = -68

y = -34

Since

x = 64 + y .

x = 64 - 34

x = 30

Hence x = 30 and y = -34

see below

step-by-step explanation:

x+2> 6

subtract 2 from each side

x+2-2 > 6-2

x > 4

there is an open circle at 4 since is it is greater than and the line goes to the right