Is #ln(x) +ln(x)# the same thing as #ln(x^2)# or #2ln(x)#????
2 Answers
Both!
Explanation:
Sum of logarithms with same base is equivalent to the logarithm of the product.
So:
It also turns out that:
So technically:
You can also just treat the
It depends, a little...
Explanation:
From basic properties of arithmetic, it is always true that:
#ln(x)+ln(x) = 2ln(x)#
Is it always true that
For the real logarithm with
Note that any non-zero complex number
#z = r(cos theta + i sin theta) = re^(itheta)#
for some real number
The suitable value of
Then:
#ln(z) = ln(re^(itheta))#
#color(white)(ln(z)) = ln(r) + ln(e^(itheta))#
#color(white)(ln(z)) = ln(r) + itheta#
#color(white)(ln(z)) = ln abs(z) + i Arg(z)#
In particular if
#ln(z) = ln(-z)+pi i#
So, for example:
#2ln(-3) = 2(ln(3)+pi i) = 2ln(3)+2pi i != 2ln(3) = ln(9) = ln((-3)^2)#
So the property
More generally, it fails for any complex number