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calculus I homework. please help!
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Find y when \[2^{x}=32^{y}, 32^{x}=16^{y+1}\]
these are my answer choices 1. y =5/21 2. y =1/7 3. y =2/21 4. y =4/21 5. y =2/7
|dw:1377841413596:dw|
\[32=2^5\\16^{y+1}=32^x=(2^5)^x=(2^x)^5=(32^y)^5=(32)^{5y}\\so\\16^{y+1}=32^{5y}\\2^{4(y+1)}=2^{10y}\\4y+4=25y\\21y=4\\y=\frac{4}{21}\]
|dw:1377841437083:dw|
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