1
$\begingroup$

Find the domain of $ \large e^{\ln\frac{x}{x+8}} $

My solution:

Set $f(x) = \large e^{\ln\frac{x}{x+8}}$, Now, we know $ e^{\ln x} = x$. So,$f(x) = \frac{x}{x+8}$

But again for the domain, since the $\ln $ cannot take negative values we have to assume that $\frac{x}{x+8} \gt 0$, which gives $ \frac{x}{x+8} \gt 0 \implies x<-8 \text{ or } x\gt 0$

but, according to someone I know it's actually, $\frac{x}{x+8}\neq 0 $.

But that mean we are operating the $e$ function first and then the rest, but why shouldn't we take care of the $\ln $ first? where exactly I am going wrong?

  • 0
    In complex analysis, they would use $\log$ for the natural logrithm, not $\ln$. Ergo, this is real.2012-02-13

4 Answers 4

1

You are right. Even though $e^{\ln \frac{x}{x+8}}=\frac{x}{x+8}$ everywhere that the left-hand side is defined, that does not mean that the left-hand side is defined when $\frac{x}{x+8}$ is negative.

Imagine that you are a simple scientific calculator. Mine puts an "E" in the display if I try to divide by $0$, or try to take the $\ln$ of a negative number, or ask it to compute $\sin^{-1}$ of a number $>1$. The calculator then refuses to go on until it is reset.

To find the largest set of reals at which an expression $\mathcal{E}(x)$ defines a function, pretend that you are a calculator, and are evaluating the expression exactly as it is written. If, for the real number $x$, at some stage of the calculation, the calculator would refuse to go on, then $\mathcal{E}(x)$ does not define a function at $x$. (I am lying a bit. My calculator also gets upset when it is asked to evaluate $2^{500}$.)

  • 0
    +1, Nice $a$nalogy, btw this is exactly how I think while doing domain problem :)2012-02-14
3

You are doing things OK. When considering the sign of $y = \frac{x}{{x + 8}}$ you have to make two systems of inequalities, namely:x > 0 \wedge x + 8 > 0 and $x < 0 \wedge x + 8 < 0$ There is no $\leq$ or $\geq$ since we don't want the $\log$ to blow up at $0$. This two inequalities will let you see $\mathbb{D} = ( - \infty , - 8)\cup (0,\infty ) $ That is, either both numerator and denominator are $+$ thus "$+\times+=+$" or both are negative, thus "$-\times-=+$"

2

You are correct; your friend is incorrect.

But, you have one bad phrasing:

\text{"since the }\ln\text{ cannot take negative values we have to assume that }\textstyle{x\over x+8}>0 ".

is not what you want to say. The $\ln$ can take negative values (for instance $\ln({1\over 1+8})$) and that would be ok; you could still take $e$ raised to that power ($e^{\ln({1\over 1+8})}= {1\over 1+8} $). The problem is that the logarithm of a nonpositive number is not defined. So, what you want to say is: \text{ "since the $\ln$ of a nonpositive number is undefined we have to assume that }\textstyle{x\over x+8}>0"


And, incidentally, it is not a good idea to start off with simplifying $e^{\ln({x\over x+8})}={x\over x+8}$. This "hides the bad points" when you just look at the right hand side; and in fact this equality is valid only then $x$ is in the domain of your function.

1

Your argument is (almost) correct. Since the domain is the reals, in order for the expression $e^{\log\left(\frac{x}{x+8} \right)}$ to be defined, we need for $ \frac{x}{x+8} > 0 $ which occurs when $x \in (-\infty, -8) \cup (0, \infty)$. Note that $x = 0$ is not allowed since $\ln(0)$ is undefined.

The second answer would be correct one was finding the domain for the complex numbers. There, $\log(z) = \ln(z) + i Arg(z)$ is defined (although one gets in to the sticky business of dealing with branche cuts) whenever $z \neq 0$. So the domain over the complex numbers is $\mathbb{C} \setminus \{0,-8\}$.

  • 0
    @bgins: Thank you. Fixed.2012-02-13