If (1/4)^x+y=256 and log(small4)(x-y)=3, calculate the exact values of x and y algebraically. Please help me out. Thank you!

since 1/4 = 4^-1 and 256 = 4^4 = 1/4^-4, you have

1/4^(x+y) = 1/4^-4
log_4(x-y) = 3 means x-y = 4^3
so now just solve
x+y = -4
x-y = 64
now solve for x and y

Did you know?

Did you know that logarithms can help solve exponential equations? In the given equation, (1/4)^x+y=256, we can use logarithms to find the values of x and y. By taking the logarithm base small4 of both sides and using the properties of logarithms, we get log(small4)(x+y)=log(small4)(256). Simplifying further, we know that log(small4)(256)=4. Therefore, we have log(small4)(x+y)=4. Now, let's look at the second equation, log(small4)(x-y)=3. By using the logarithmic property that log(a) - log(b) = log(a/b), we can rewrite this equation as log(small4)(x/y)=3. Since we found earlier that log(small4)(256)=4, we can rewrite the equation as log(small4)(256/x)=3. Evaluating this equation, we raise small4 to the power of 3, which gives us 64. Thus, we have 256/x=64. By solving for x, we find that x=4. Finally, by substituting the value of x into the first equation, we can solve for y.