Question

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

Answer 1

Sticking with Real logarithms, this expands as:

`ln(sqrt(a)(b^2+c^2)) = 1/2 ln(a) + ln(b^2+c^2)`

Explanation:

If `x, y > 0` then `ln(xy) = ln(x)+ln(y)`

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!=0` or `c!=0`, resulting in a strictly positive value for `b^2+c^2`.

Also note that if `a > 0` then `ln(sqrt(a)) = ln(a^(1/2)) = 1/2 ln(a)`

Hence:

`ln(sqrt(a)(b^2+c^2))`

`=ln(sqrt(a))+ln(b^2+c^2)`

`=1/2 ln(a) + ln (b^2+c^2)`

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

`=1/2 ln(a) + ln (b+c i) + ln (b - c i)`

based on the fact that `b^2+c^2 = (b+c i)(b - c i)`, but there are some problems with this.

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

`0 = ln(1) = ln(b^2+c^2) != ln(b+ci) + ln(b-ci) = ln(-1)+ln(-1) = 2 pi i`

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