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

1 Answer
George C.
Dec 12, 2015

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.

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