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8^(y - 2) = 2^y a.y = 1 b.y = 3 c.y = -3 d.y = -1
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Can you convert \( 8^{y-2}\) to a power of 2? That is \( \large2^?\) ?
im lost on what to do
Okay. Let me make it bigger \[ \Large8^{y-2} \] is on the left. \[ \Large 2^y \] is on the right. You can solve this problem if you can express both terms as the same base number to some power. So how can you convert \[ \Large8^{y-2} \] to \[ \Large2^{something} \]
do you divide it by 4?
Here's a hint \[ \Large 8 = 2^3 \]
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so the answer would be 3?
\[\Large8^{y-2} \] is what you are trying to convert. I just told you \[\Large 8 = 2^3\] So what can you do?
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