if $8^{y}=16^{y + 2}$, what is the value of $y$?\n-8\n-4\n-2\n-1

Explanation:

Step1: Rewrite bases as powers of 2

Rewrite $8^y$ as $(2^3)^y = 2^{3y}$ and $16^{y + 2}$ as $(2^4)^{y+2}=2^{4(y + 2)}$. So the equation becomes $2^{3y}=2^{4(y + 2)}$.

Step2: Set exponents equal

Since the bases are the same, we can set the exponents equal: $3y=4(y + 2)$.

Step3: Expand and solve for y

Expand the right - hand side: $3y = 4y+8$. Subtract $3y$ from both sides: $0=y + 8$. Then subtract 8 from both sides to get $y=-8$.

Answer:

-8