If 2^{3y+x}=16 and 2^{2y−x}=64

$2^{3y} + x = 16$ and $2^{2y} - x = 64$

We need to find the value of $y$.

Step 1: Express $x$ from both equations

From the first equation: $x = 16 - 2^{3y}$

From the second equation: $x = 2^{2y} - 64$

Step 2: Equate the two expressions for $x$

$16 - 2^{3y} = 2^{2y} - 64$

$2^{3y} + 2^{2y} = 16 + 64 = 80$

Step 3: Factorize

$2^{2y} (2^y + 1) = 80$

Let $2^y = t$, then $2^{2y} = t^2$, so

$t^2 (t + 1) = 80$

$t^3 + t^2 - 80 = 0$

Step 4: Solve cubic equation

Try integer factors of 80. If $t = 4$:

$4^3 + 4^2 - 80 = 64 + 16 - 80 = 0 ✅$

So, $2^y = 4$

Step 5: Solve for $y$

$2^y = 4 = 2^2$

$y = 2$

✅ Answer: $y = 2$