How do we solve for x and y where y=(1/3)log2(x) and 8^(2y-1) = 2(x-4)

From the 1st one,I got x = 2^(3y) & by substituting it in the 2nd resulted 8^(2y-1) = 2((2)^(3y) - 4)

How to proceed?

ok so far, but messy

Check if the 2nd equation wasn't
8^(2y-1) = 2^(x-4)

Thanks Reiny I solved it

To solve the equation, 8^(2y-1) = 2((2)^(3y) - 4), you can follow these steps:

Step 1: Simplify the right side of the equation:
2((2)^(3y) - 4) = 2(8^y - 4)
= 2(8^y) - 2(4)
= 16^y - 8

Step 2: Rewrite 8 as 2^3, and use the exponent properties:
16^y = (2^4)^y = 2^(4y)

So, the equation becomes:
2^(4y) - 8 = 8^(2y-1)

Step 3: Rewrite 8 as 2^3:
2^(4y) - 2^3 = (2^3)^(2y-1)

Step 4: Apply the exponent properties to simplify further:
2^(4y) - 8 = 2^(6y - 3)

Step 5: Now, we can set the exponents equal to each other since the bases are the same:
4y = 6y - 3

Step 6: Rearrange the equation to isolate y:
4y - 6y = -3
-2y = -3
y = (-3) / (-2)
y = 3 / 2
y = 1.5

Step 7: Substitute the value of y back into the original equation to solve for x:
x = 2^(3y)
x = 2^(3 * 1.5)
x = 2^(4.5)
x = √(2^9)
x = 2^4
x = 16

So, the solution to the equations is x = 16 and y = 1.5.