Find x when 2^x=16^y, 16^x=8^(y-1)
Okay when I exponentiate for the first part I get y=x/3 and for the second part I get y=4x-1/3. However, I don't understand how to get one x. I don't know what the next step is. Please help me.

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\[2^x=16^y\]
\[2^x=2^{4y}\]
\[x=4y\]

then \[16^x=8^{y-1}\]
\[2^{4x}=2^{3(y-1)}\]
\[4x=3y-3\]

should be good from there yes?

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