Question

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

Answer 1

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)`.

Answer 2

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)`.