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

Since $8 = 2^3$ and $16=2^4$, we can rewrite the equation $8^y = 16^{y + 2}$ as $(2^3)^y=(2^4)^{y + 2}$.

Step2: Apply power - of - a - power rule

The power - of - a - power rule $(a^m)^n=a^{mn}$. So, $(2^3)^y = 2^{3y}$ and $(2^4)^{y + 2}=2^{4(y + 2)}=2^{4y+8}$. The equation becomes $2^{3y}=2^{4y + 8}$.

Step3: Set exponents equal

If $a^m=a^n$, then $m = n$ for $a>0,a\neq1$. So, $3y=4y + 8$.

Step4: Solve for y

Subtract $3y$ from both sides: $0=y + 8$. Then subtract 8 from both sides to get $y=-8$.

Answer:

-8