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