# simplify log

• Nov 16th 2007, 12:52 AM
BruceBronson
simplify log
This isnt really homework, its a problem im attempting

http://img405.imageshack.us/img405/9469/93511342nl4.jpg

IS THIS SOLUTION CORRECT?

for the first equation i get C_2 = Log((1-a)/(1+a)) I Do NOT see how they can get that answer

then this gives the solution for the second bit as

C_1 = ( log ((1+b)/(1-b) + log ((1-a)/(1+a)) )^-1

HELP PLEASE, thank you
• Nov 16th 2007, 01:20 AM
earboth
Quote:

Originally Posted by BruceBronson

...

for the first equation i get C_2 = Log((1-a)/(1+a)) I Do NOT see how they can get that answer

...

Hello,

you are supposed to know that:

$\displaystyle -\ln\left(\frac ab \right) = \ln(1)-\ln\left(\frac ab \right) = \ln\left(\frac{1}{\frac ab} \right) = \ln\left( \frac ba \right)$

Now consider your first equation:

1. Divide by $\displaystyle C_1$
2. Subtract the ln-value:

$\displaystyle -\ln\left(\frac{1+a}{1-a}\right)=C_2$ Use the property I mentioned above and you'll get

$\displaystyle \ln\left(\frac{1-a}{1+a}\right)=C_2$