# How do you expand ln ((sqrt(a)(b^2 +c^2))?

Dec 12, 2015

Sticking with Real logarithms, this expands as:

$\ln \left(\sqrt{a} \left({b}^{2} + {c}^{2}\right)\right) = \frac{1}{2} \ln \left(a\right) + \ln \left({b}^{2} + {c}^{2}\right)$

#### Explanation:

If $x , y > 0$ then $\ln \left(x y\right) = \ln \left(x\right) + \ln \left(y\right)$

Assuming we're dealing with Real values here and everything is well defined, we must have $a > 0$ and ${b}^{2} + {c}^{2} > 0$. That is, at least one of $b \ne 0$ or $c \ne 0$, resulting in a strictly positive value for ${b}^{2} + {c}^{2}$.

Also note that if $a > 0$ then $\ln \left(\sqrt{a}\right) = \ln \left({a}^{\frac{1}{2}}\right) = \frac{1}{2} \ln \left(a\right)$

Hence:

$\ln \left(\sqrt{a} \left({b}^{2} + {c}^{2}\right)\right)$

$= \ln \left(\sqrt{a}\right) + \ln \left({b}^{2} + {c}^{2}\right)$

$= \frac{1}{2} \ln \left(a\right) + \ln \left({b}^{2} + {c}^{2}\right)$

If we allow Complex logarithms, then we might try to say something like:

$= \frac{1}{2} \ln \left(a\right) + \ln \left(b + c i\right) + \ln \left(b - c i\right)$

based on the fact that ${b}^{2} + {c}^{2} = \left(b + c i\right) \left(b - c i\right)$, but there are some problems with this.

For example, if $b = - 1$ and $c = 0$ then we find:

$0 = \ln \left(1\right) = \ln \left({b}^{2} + {c}^{2}\right) \ne \ln \left(b + c i\right) + \ln \left(b - c i\right) = \ln \left(- 1\right) + \ln \left(- 1\right) = 2 \pi i$

So this Complex identity does not quite work and is messy to fix.