# Core 3 help

1. Jun 24, 2007

### CathyLou

Hi.

Could someone please help me with the following questions? I'm totally stuck so any feedback would be really appreciated. I'll post the questions one at a time.

Express as single log. functions:

(i) ln (x + 1) - 3 ln (1 - x) + 2 ln x

I got up to ln ( (x + 1) / (1 - x)^3 ) / ln x^2 but I don't know what to do next or whether this is even correct.

Thank you.

Cathy

2. Jun 24, 2007

### Staff: Mentor

Using the identity ln(xy) = ln x + ln y, one can extend it to ln (xyz) = ln x + ln y + ln z, and ln (x/y) = ln x - ln y.

So in the problem ln (x + 1) - 3 ln (1 - x) + 2 ln x

one obtains ln [(x+1)/(1-x)3] + ln x2 which can be further simplified by bringing x2 inside the logarithm operation.

ln [(x+1)x2/(1-x)3]

Reference: http://mathworld.wolfram.com/Logarithm.html

3. Jun 24, 2007

### CathyLou

Thanks so much for your help.

Cathy

4. Jun 24, 2007

### CathyLou

Could someone please give me a tip for how to answer this one as I've no idea what to do? I'd really appreciate it.

Given ln (xy^3) = m and ln (x(^3)y(^2)) = n, find ln root(xy) in terms of m and n.

Cathy

5. Jun 24, 2007

### Staff: Mentor

By root (xy), does one mean $\sqrt{xy}$?

If so, then ln (xy)1/2 = 1/2 ln xy = 1/2 (ln x + ln y)

and one also needs to exand the equations ln (xy3) = ln x + 3 ln y = m, and similarly for the other equation, then rearrage to x and y in terms of m and n.

Last edited: Jun 24, 2007
6. Jun 24, 2007

### CathyLou

Yeah, that's what I meant. I just don't understand how to connect that to m and n.

7. Jun 24, 2007

### CathyLou

Oh, I've got that one now.

I have one last query and that is how to draw the graphs (and asymptotes) of y = 2 + ln x and y = - ln (x - 3).

Thank you.

Cathy

8. Jun 24, 2007

### Staff: Mentor

Well, certain as x gets very large, y = (2 + ln x) ~ ln x,

and similarly as x gets very large, i.e. x >> a, then x+a ~ x.

Also, think of the range for ln (3-x).