Question

PLEASE SOLVE FOLLOWING FOR Y: LN(Y+1) + LN(Y-1)=2X+LNX (From Unit 1 Problem Set 2 1H.3A)

I don't understand why after taking e on both sides the terms are multiplied instead of added, according to the solutions. For the first part of the equation, I could possibly see how it is the inverse of ln(x*y)=lnx + lny. I don't see that for the right side of the equation though.

Answer 1

ln(y+1) +ln(y-1) = 2x +ln(x) ; the left rules out as: ln(a)+ln(b)=ln(ab)

ln[(y+1)(y-1)] = 2x +ln(x); subtract ln(x) from both sides

ln[(y+1)(y-1)] - ln(x) = 2x ; left rules as: ln(a)-ln(b)=ln(a/b)

ln[(y+1)(y-1)/x] = 2x ; now "e" both sides

(y+1)(y-1)/x = e^(2x) ; e^(2x)

(y+1)(y-1) = x e^(2x) ... which is definantly the long way around it
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start over like this:

ln(y+1) +ln(y-1) = 2x +ln(x) ; now "e" both sides to get..

e^(ln(y+1) +ln(y-1)) = e^(2x +ln(x)) ; b^(m+n) = b^(m)*b^(n)

e^(ln(y+1)) * e^(ln(y-1)) = e^(2x) * e^(ln(x)) ; simplify

(y+1)(y-1) = e^(2x) x

y^2 -1 = x.e^(2x)

y^2 = x.e^(2x) + 1

y = +- sqrt(x.e^(2x) +1)

I believe thats what we get to ...

Answer 2

Thank you. I think I see it now, I wasn't applying b^(m+n)=b^(m)*b^(n). I didn't see 2x and lnx as as (m+n), but as separate.

Answer 3

LN(Y+1) + LN(Y-1)=2X+LNX

LN[(Y+1)*(Y-1)]=LN(e^{2X})+LN(X) : LN(A)+LN(B)=LN(A*B); A=LN(e^{A})

LN(Y^{2}-1) =LN[X*e^{2X} ]

Y^{2} -1=X*e^{2X}

Y^{2}=X*e^{2X}-1

Y=+/-sqrt(X*e^{2X}-1)