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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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Hi, Welcome to math.SE. Please try to use $\LaTeX$ here. If $C$ is a constant, it is better to cancel $\frac 12$ from both sides.. and then use $\ln a - \ln b=\ln \frac ab$ –  TheJoker Nov 15 '12 at 9:24
    
how did $\ln C$ become $C$? –  user31280 Nov 15 '12 at 9:27
2  
What about $\ln x - \ln y = \ln (x/y),\ \ln x + \ln y = \ln (xy)$ and $\ln |C| = \frac12\ln (C^2)$? –  user1551 Nov 15 '12 at 9:27
    
@TheJoker so that becomes, $$\ln|y+1| - \ln|y-1| = \ln|x+1| - \ln|x-1|$$ can it be written as $$\ln|(y+1)-(y-1)| = ln|(x+1)-(x-1)|$$ –  TPSstar Nov 15 '12 at 9:49
    
@F'OlaYinka Sorry, would you please help to let me know what is $$\ln C $$ –  TPSstar Nov 15 '12 at 9:54

1 Answer 1

up vote 1 down vote accepted

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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Superb! Thanks :) –  TPSstar Nov 15 '12 at 11:32
    
if you are satisfied, please consider accepting the answer. meta.math.stackexchange.com/questions/3399/… –  TheJoker Nov 15 '12 at 11:42
    
$$ \ln |\frac {y+1}{y-1}*C| = \ln | \frac {x+1}{x-1}| $$ if $$ x=2 , y=2 $$ How can i find value of C? do i need to apply log rules again? –  TPSstar Nov 15 '12 at 12:33
    
putting the values, we get,$\ln 3C = \ln 3$.Take out the log, then we get $3C=3 \iff C=1$. –  TheJoker Nov 15 '12 at 12:37
    
Thank you, you are awesome :) –  TPSstar Nov 15 '12 at 12:38

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