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until the integrating parts is correct:
ln(y-1) = ln x + B
where B is an arbitrary constant
pls explain why is it so, for the following solution.
On solving this eqn for y, by first taking the exponential of both sides, we obtain
y = 1 ± e^lnx+B
= 1 ± e^B . e^lnx
= ???
algebra:
\[y-1 = x e^{C}\] (let's just call \[e^{C}\] k since its just a constant anyway)
\[y = kx+1\]
I think the focus of your question was on the step of getting from \[e^{ln(x) + C}\] to \[x e^{C}\]. It is a property of exponents that \[n^{a+b} = n^a n^b.\]
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