If (1/4)^𝑥+𝑦 = 256 and 𝑙𝑜𝑔4(𝑥 − 𝑦) = 3, calculate the exact values of 𝑥 and 𝑦 algebraically.

Answer 1

(1/4)^x = 4^-x
So, assuming the usual carelessness with parentheses, you must have meant
(1/4)^(x+y) = 256
4^-(x+y) = 4^4
-(x+y) = 4

log4(x-y) = 3
x-4 = 4^3 = 64
so now just solve
x+y = -4
x-y = 64
x = 30
y = -34