Question

Original question. e^(log(x/(x+8))) Find domain.

e^(log(x)-log(x+8)), (e^log(x))/(e^log(x+8))
then
(e^(ln(x)/ln(10)))/(e^(ln(x+8)/ln(10)))
then
(x^(1/ln(10)))/((x+8)^(1/ln(10)))

Now, the bottom obviously cannot equal -8. We know that 1/ln(10)∉ℚ but 1/ln(10)∈ℝ which leads me to the conclusion that 1/ln(10) is a transcendental number. Now, I think that any real x of 0 or above can be raised to a transcendental and be in the real set, such that x^n∈ℝ for x≥0. However, for x<0, I'm thinking that ∀(x,n)P(x^n∈ℂ) for all n as a member of the transcendental set.

If this is so--then x>0 is

Answer 1

ln(a)/ln(10) != ln(a/ln(10)), that's where you went wrogn with your proof, I think

Answer 2

ln(a)/ln(10) = ln(a^(1/ln(10)))

Answer 3

well i don't know about this application, i was just using this ab/c=a(b/c) , i see it doesn't apply to logs

Answer 4

Yeah, so if ∀(x,n)P(x^n∈ℂ) when n is a transcendental and x<0, then x>0 is teh domain. But if it isn't, then there might be a different domain

Answer 5

a*ln(b)=ln(b^a)

Answer 6

Someone actually good at math should take a look at this. :( I don't know if I'm doing stuff correctly

Answer 7

you're probably right, i'm lost, will show this to ffm later when he's on

Answer 8

If you want a decent answer right now, I'm going to have to go with x>0. I'm almost certain of this, unless I made a worse mistake earlier.

Answer 9

For \(f(x)=e^x\), \(D(f(x))=\{x: x\in \mathbb{R}\}\). For \(g(x)=\log_{10}(x)\), \(D(g(x))=\{x: x>0\}.\)
\(D((f\circ g)(x))=\{x: x\in \mathbb{R} \wedge x > 0\}=\{x: x>0\}.\)
If \(x=\frac{x'}{x'+8}\), then \(D(f \circ g)(x)=\{x: x>0 \}\) since \(\frac{x'}{x'+8}>0\) \(\forall x'>0.\)

Four months late, but you were right.