Use the properties of logarithms to expand the logarithmic function

ln[(x^2+1)(x-1)]

Would this be the correct answer?
ln[x^3-x^2+x-1]
I can't seem to expand it any further than that.

your expansion is 100% correct

you can just say

lnx^3-lnx^2+lnx-ln1
lnx^3-lnx^2+lnx-0
which is
lnx^3-lnx^2+lnx

no, you cannot.

ln(x) + ln(y) = ln(xy)

ln(x+y) is NOT ln(x) + ln(y)

instead of expanding the polynomial, you could do

ln[(x^2+1)(x-1)]
= ln(x^2+1) + ln(x-1)

To expand the given logarithmic function, we can use the properties of logarithms. Let's break down the steps:

1. Start with the given logarithmic expression: ln[(x^2+1)(x-1)].

2. Apply the property of logarithms, which states that the logarithm of the product of two numbers is the sum of their logarithms. So, we can rewrite the expression as: ln(x^2+1) + ln(x-1).

3. Now, let's focus on simplifying ln(x^2+1):
- The property of logarithms that says ln(ab) = ln(a) + ln(b) can also be used to rewrite ln(x^2+1) in a more thorough way.
- The term ln(x^2+1) can be expressed as: ln(x^2) + ln(1) since 1 is just a constant and ln(1) = 0.
- So, we get ln(x^2) + ln(1) = 2ln(x) + 0 = 2ln(x).

4. Putting it all together, we have: ln(x^2+1) + ln(x-1) = 2ln(x) + ln(x-1).

Therefore, the expanded form of ln[(x^2+1)(x-1)] is 2ln(x) + ln(x-1).
The answer you provided, ln[x^3-x^2+x-1], is incorrect.