How do you simplify this and find the domain?
e^logx/(x + 8)

Question

How do you simplify this and find the domain?
e^logx/(x + 8)

anonymous · Answer

$$\frac{e ^{\log(x)}}{x+8}?$$

anonymous · Answer

no the x +  8 where you have the x beside log

anonymous · Answer

use the equation on the bottom rewrite

sasogeek · Answer

$$\huge e^{{log\frac{x}{x+8}}} $$

sasogeek · Answer

is that the question?

anonymous · Answer

yes

campbell_st · Answer

well e^log(x) = x so in your question it simplifies to x/(x+8) 

as far ad the domain is concerns x cannot equal 8... but can be any other real values...

so domain, all real x except x=8

sasogeek · Answer

cambell, $$\huge e^{log(x)} \neq x  \ but \ rather, e^{ln(x)}=e^{\log _{e}x=x}$$ \]

campbell_st · Answer

well use change of base to change it to base e logs

the domain is still unchanged.

campbell_st · Answer

so ln (x/x+8) = ln(x/x+8)/ln10  thats change of base...

campbell_st · Answer

then the problem is e^ln(etc)...

campbell_st · Answer

simplfies to e^ln(x/10(x+8) = x/10(x+8) domain is unchanged.

sasogeek · Answer

$$\huge \text {taking only the exponent...} $$$$\huge log(\frac{x}{x+8})=\frac{ln(\frac{x}{x+8})}{ln(10)}=$$ ....

sasogeek · Answer

how is it $$\huge \frac{x}{10(x+8)} ? $$

campbell_st · Answer

well log (x/(x+8) -10) then the domain is all real x except x cannot equal 2...

sasogeek · Answer

well the questions says simplify... not solve... :P

campbell_st · Answer

it says state the domain

anonymous · Answer

logx/x + 8 = Inx/(x + 8)In10?

sasogeek · Answer

change of base

anonymous · Answer

log(4)/log(e) != 4/e

anonymous · Answer

Can't cancel out functional operators as with dviison.

sasogeek · Answer

well i'm lost here so i'll delete my solutions, if i find something i can settle with as a solution and i'm sure of it, i'll let u guys know, at the moment, i'm still confused about all the current answers but if that is the right solution, awesome! i'm the confused one here :)

anonymous · Answer

The simplification I have is (x^(1/ln(10)))/((x+8)^(1/ln(10)))

anonymous · Answer

What the hell is it with transcendental numbers.

anonymous · Answer

Alright, because the exponent is a transcendental number, I'm going to say that x must be x>0 for the top. For the bottom, (x+8) must be (x+8)>0, or x>-8.

sasogeek · Answer

$$\huge e^{ln\frac {(\frac{x}{x+8})}{ln10}}=\frac{(ln10)x}{x+8} $$

anonymous · Answer

Continued discussion : http://openstudy.com/study#/updates/4f38ae13e4b0fc0c1a0dda3b