Question

Expand using properties of logs:
ln X^2(X^2-1)/√X+1

Answer 1

All of that stuff is inside of one log?

Answer 2

Its a quotient problem. The x^2(x^2-1) in the top part of the quotient and the √x-1 is under

Answer 3

This one is a bit of a doozy, lot of little steps.

Answer 4

A rule of logs states:

log(a/b) = log(a) - log(b)

So ln x^2(x^2-1)/√x+1
becomes,

ln x^2(x^2-1) - ln√x+1

Answer 5

Understand how I applied that step?
~Separating the numerator and denominator into different logs.

Answer 6

yes i'm there

Answer 7

Another important rule of logs:

log(b*c) = log(b) + log(c)

When we multiply we can split them up into the sum of logs.
Understand how we might apply this rule? :o

Answer 8

how are you applying it

Answer 9

In the first log, we have:

ln x^2*(x^2-1)

See the multiplication?
we can rewrite this as 2 logs being added together.

Answer 10

so x^2+x^2+x^2-1?

Answer 11

Hmm I'm not sure what you did there.. we don't want to multiply it out.
We want to apply the log rule:

ln x^2(x^2-1) - ln√x+1 = ln x^2+ ln(x^2-1) - ln√x+1

Answer 12

There's a bunch more little steps from here.

Answer 13

i understand what you did where do you go from here?

Answer 14

my next step was 2lnx+2lnx-1-1/2lnx+1

Answer 15

Your first term looks good.
2lnx

Your last term also looks good.
(1/2)ln(x+1)

You rewrote the root as a 1/2 power, and pulled it outside.

Let's fix up the middle term.

Answer 16

ln(x^2-1)

We have the `difference of squares` inside of the log.
We can factor the difference of squares into conjugates.

(x^2-1) = (x-1)(x+1)

So that gives us:
ln(x^2-1) = ln(x-1)(x+1)

All of that inside of the log.

Answer 17

Then you might notice that again, we have multiplication.
So we can write this as the sum of logs:

ln(x-1) + ln(x+1)

That would be our middle term.

Answer 18

ok so like foil method.i got i jeeez!

Answer 19

*it