Question

FFM: simplify and find the domain. \[\huge e^{log\frac{x}{x+8}} \] please show your working :) thanks, I'm confused about my answers Original question: http://openstudy.com/study#/updates/4f38ae13e4b0fc0c1a0dda3b

Answer 1

I think you mean \( \ln \)?

Answer 2

well in the original question, it says \(\large log\frac{x}{x+8} \)

Answer 3

hm but \( \log \) in general means the common logarithm and it depends on the context of usage. For example in computer science \( \log \) would suggest base 2 logarithm, in algebra it's base 10 and in calculus it's base \( e\), which is again the natural logarithm.

Answer 4

so then \(\ log \) here could suggest log in terms of calculus... which would imply ln?

Answer 5

Yes, here it should be interpreted as natural logarithm.

Answer 6

I took log to be base 10, not ln. My bad.

Answer 7

which means it simplifies to \(\large \frac{x}{x+8} \)

Answer 8

Yup.

Answer 9

The domain for that should be straightforward.

Answer 10

Coming back to the question,

>> simplify and find the domain: \( \large e^{\ln\frac{x}{x+8}} \)

Lets say f(x) = \( \large e^{\ln\frac{x}{x+8}} \)

Now, we know \(e^{\ln x} = x \)

So,\( f(x) = \frac{x}{x+8} \) is the simplified version

For the domain calculation we have to note that \( \frac{x}{x+8} \ge 0\) since \(\ln\) of a negative value is not define in reals.

So \( \frac{x}{x+8} \ge 0 \implies x<-8 \text{ or } x\geq 0 \) and this should also be the domain of our f(x) itself .

Answer 11

Whoa, forgot that last part, where the ln of a negative isn't real. Thanks for reminding.

Answer 12

Glad to help :)

Answer 13

actually, \[\frac{x}{x+8}\neq 0\]

Answer 14

nik, I think I can see your point now, I will work on it a bit more and will be back on this.

Answer 15

Actually it should be \( \frac{x}{x+8} \gt 0 \). I had made a mistake of not taking care of x=0 case which is undefined.

so the domain is \( x \in (-\infty, -8) \cup (0, \infty) \)

Answer 16

Please note I am assuming only real.