# Manipulating natural log

1. Dec 12, 2015

### Sturk200

How is it true that:

Log[L+(Z^2+L^2)^(1/2)] - Log[-L+(Z^2+L^2)^(1/2)] = 2{Log[L+(Z^2+L^2)^(1/2)] - Log[Z]}

?

2. Dec 12, 2015

### Staff: Mentor

Is this homework?

The equality is not very complicated to check. Use the properties of logarithms to get to the form log(x) = log(y), and then check if x = y.

3. Dec 12, 2015

### Sturk200

Not homework, but part of a textbook problem that I'm using to study. I can get as far as the form log(x) = log(y) by turning the differences into quotients and the multiplicative prefactor on the right into an exponent. I guess I'm having trouble with the easiest part, seeing the algebraic identity.

4. Dec 12, 2015

### SteamKing

Staff Emeritus

$log\; [L + \sqrt{(Z^2 + L^2)}] -log\; [-L + \sqrt{(Z^2 + L^2)}]=2log\; [L + \sqrt{(Z^2 + L^2)}] - log (Z)$ ?

The Laws of Logarithms are:

$log\; (a) - log\; (b) = log\;(\frac{a}{b})$

$log\; (a+b) - log\; (a-b) = log\;(\frac{a+b}{a-b})$

$log\;(a^b) = b\;log\;(a)$

$log\;[(a+b)^c]=c\;log\;(a+b)$

5. Dec 12, 2015

### Sturk200

Thanks for your reply. Yes, I understand this much. So the problem becomes showing that

(L + Sqrt[Z^2 + L^2])/(-L + Sqrt[Z^2 + L^2]) = [(L + Sqrt[Z^2 + L^2])/Z]^2

Maybe this is me being dumb, but I don't know how to get from the left side to the right side.

6. Dec 12, 2015

### nrqed

If you start from the left side, simply multiply by $(L+\sqrt{Z^2+L^2})/(L+\sqrt{Z^2+L^2})$. This is the same trick as when we multiply by conjugates of complex numbers to get rid of a term (a+ib) in the denominator.

7. Dec 12, 2015

### Sturk200

My lord, that is simple. Thank you.