Homework Help: 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

Astronuc

Staff Emeritus
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

Astronuc

Staff Emeritus
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

Astronuc

Staff Emeritus
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).