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How to simplify this natural logarithm $\cfrac12\ln|y+1|-\cfrac12\ln|y-1|+\ln|C| =\cfrac12\ln|x+1|-\cfrac12\ln|x-1|$

if I apply the logarithm rule

$\ln\sqrt{|y+1|} - \ln\sqrt{|y-1|} + \frac12\ln (C^2) = \ln\sqrt{|x+1|} -\ln\sqrt{|x-1|}$

Please help further..

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    no.. i'll write the answer as it has bypassed the *comment stage*..2012-11-15

1 Answers 1

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First, Note that $x,y \neq \pm1$ and $\ln C = \frac 12 \ln C^2$. Denote $C^2=k >0$. So the equation becomes

$\frac 12 (\ln |y+1| -\ln |y-1| + \ln k ) = \frac 12 (\ln |x+1| - \ln |x-1| )$

Cancelling $\frac 12$ and using basic logarithm identities, we get,

$ \ln \left ({ k \left |\frac {y+1}{y-1}\right |}\right ) = \ln \left ( \left | \frac {x+1}{x-1} \right | \right ) $

Now, we get, $k\left |\frac {y+1}{y-1}\right | = \left | \frac {x+1}{x-1} \right |$

Now simplify it. Best way is to break into different cases.

Case 1: $ |y| > 1 $ and $|x| >1$

Gives $k \frac {y+1}{y-1} = \frac {x+1}{x-1} $ Solve for $y$ in term of $x$ or vice versa as desired.

Other cases can be dealt with similarily.

Note: Seems like you are starting to learn about logarithms, so I'll write the facts I used;

$\ln a + \ln b = \ln ab$ $\ln a - \ln b = \ln \frac ab$ $\ln a = \ln b \iff a=b$

all of these holds for $a,b \in \mathbb R^+$

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    nope, you haven't seen real awesome ppl here yet.. welcome again here and have a nice time..2012-11-15