Use the properties of logaritms to expand the expression

[pooker]

Homework Statement

IN * ((4x^5 - x -1)(square root x-7)) / (x^2 + 1)^3

The Attempt at a Solution

in(4x^5 - x - 1) + in(square root x-7) - in(x^2 + 1)^3

in(4x^5 - x - 1) + 1/2in(x-7) - 3in(x^2 +1)

5in4x - inx - 1 + 1/2(inx -in7) - 3 (2inx + in)


I know that's not right but its all I can think of at the moment.

 

[Mark44]

Homework Statement

IN * ((4x^5 - x -1)(square root x-7)) / (x^2 + 1)^3

The Attempt at a Solution

in(4x^5 - x - 1) + in(square root x-7) - in(x^2 + 1)^3

in(4x^5 - x - 1) + 1/2in(x-7) - 3in(x^2 +1)

5in4x - inx - 1 + 1/2(inx -in7) - 3 (2inx + in)


I know that's not right but its all I can think of at the moment.

First off, it's not in, it's ln (ell en), short for logarithm naturalis or something close to that. Before being able to help you out, can you confirm that this is the problem?
$$ln \frac{(4x^5 - x - 1)(\sqrt{x - 7})}{(x^2 + 1)^3}$$

If so, then an expression of the form ln[(AB)/C] can be rewritten as ln A + ln B - ln C.
Then because the individual expressions B and C have exponents, you can use the property of logarithms that ln a^b = b ln a.

Mistakes in your work:
1. You start with IN *. I've already mentioned that it is LN, but the mistake here is the mulitiplication symbol. ln is a function, not a number, so it's meaningless to think of using it to multiply.
2. In your later work you have "3in(x^2 +1)" and in the next line you have "3 (2inx + in)". This suggests to me that you think that "in" is a number. It's not. 2 inx + in is meaningless.