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

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

answer: 12/52

step-by-step explanation: my reasoning is that there are 12 face cards in a deck and a total of 52 cards in all.