is $y=e^{\ln(x)}$ a function??? 
Is $y=e^{\ln(x)}$ a function?

I am not sure whether this is a function because it should be equal to $y=x$ but $x$ cannot be zero so I am confused  as to wether this is a function or not
 A: A function needs a domain and a codomain, so the answer is neither yes nor no. If you add that the domain is $(0,\infty)$ and that the codomain is $\mathbb R$, then your expression defines a function. If both the domain and the codomain are equal to $\mathbb R$, then, no, it doesn't define a function.
A: Hint: We have $\exp: {\Bbb R}\rightarrow{\Bbb R}_{>0}:x\mapsto e^x$ and $\log: {\Bbb R}_{>0}\rightarrow{\Bbb R}:x\mapsto \ln x.$
These functions are inverse to each other. Can you finish?
FINISHING:
$\exp(\ln(x))=x$ for each $x\in {\Bbb R}_{>0}$ and $\ln(\exp(x))=x$ for each $x\in{\Bbb R}$.
A: It is a function with domain $(0, \infty)$ (since $\ln(x)$ has domain $(0, \infty)$). On this domain, $e^{\ln(x)}=x$.
A: Assuming you only want to work with real numbers, the natural logarithm has domain $\Bbb R^+$ and range (or codomain, whichever term you prefer) $\Bbb R$, while the exponential has domain $\Bbb R$ and range $\Bbb R^+$. Therefore, $\exp\ln x$ has maximal domain and range both equal to $\Bbb R^+$, for which it reduces to the identity function. Choose any $S\subseteq\Bbb R^+$ to be both the domain and range; then the function is the identity thereof.
A: It is the exponential function composed with the logarithm function. Since the latter operates only on positive real numbers $x,$ and the former is defined for all $\log x$ (every real number is the $\log$ of some positive number), then your expression uniquely defines a real number for each positive $x.$ In this case it represents a function. Indeed, the identity (not the constant $1$ as you think), since $\log$ and $\exp$ are inverses.
