simplify log

Quote:

Originally Posted by BruceBronson

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

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Hello,

you are supposed to know that:

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

Now consider your first equation:

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

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

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