Question

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

Explanation:

Step1: Express bases as powers of 2

Rewrite \(8^y\) and \(16^{y + 2}\) with base 2. We know that \(8 = 2^3\) and \(16 = 2^4\). So, \(8^y=(2^3)^y\) and \(16^{y + 2}=(2^4)^{y + 2}\).
Using the exponent rule \((a^m)^n=a^{mn}\), we get \((2^3)^y = 2^{3y}\) and \((2^4)^{y + 2}=2^{4(y + 2)}\).
Now the equation \(8^y = 16^{y + 2}\) becomes \(2^{3y}=2^{4(y + 2)}\).

Step2: Set exponents equal

Since the bases are the same and the equation holds, the exponents must be equal. So we set up the equation \(3y=4(y + 2)\).

Step3: Solve for y

Expand the right - hand side: \(3y = 4y+8\).
Subtract \(4y\) from both sides: \(3y-4y=4y + 8-4y\), which simplifies to \(-y = 8\).
Multiply both sides by \(- 1\) to get \(y=-8\).

Answer:

\(-8\) (corresponding to the option "−8")