Is #ln(x) +ln(x)# the same thing as #ln(x^2)# or #2ln(x)#????

2 Answers
NJ
Apr 3, 2018

Both!

#ln(x) + ln(x) = ln(x^2) = 2ln(x)#

Explanation:

Sum of logarithms with same base is equivalent to the logarithm of the product.

So: #ln(x) + ln(x) = ln(x^2)#

It also turns out that:

#ln(x^2) = 2ln(x)#

So technically:

#ln(x) + ln(x) = ln(x^2) = 2ln(x)#

You can also just treat the #ln(x)# like it's a variable and sum them. If you have #ln(x) + ln(x)#, then you have #2ln(x)#.

George C.
Apr 3, 2018

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 #2ln(x) = ln(x^2)# ?

For the real logarithm with #x > 0# the answer is yes, but the same is not true if the definition of logarithms is extended to negative and complex numbers.

Note that any non-zero complex number #z# can be represented as:

#z = r(cos theta + i sin theta) = re^(itheta)#

for some real number #r > 0# and #theta in (-pi, pi]#

The suitable value of #theta in (-pi, pi]# is #Arg(z)#

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 #z < 0# is a negative real number then:

#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 #ln(z^2) = 2ln(z)# fails for negative numbers.

More generally, it fails for any complex number #z# with #Arg(z) in (pi/2, pi]# or #Arg(z) in [-pi/2, -pi)#.

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