Prove $f(x) = 3^x$ is not onto. Give a counterexample to prove $f(x) = 3^x$ is not onto.  
A function is onto if for all $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$.  Essentially, the range of our function $f$ is equal to the codomain.  
I know that $f(x)=3^x$ is not onto, but I'm having trouble finding a counterexample to prove this.  
We know the domain is the real numbers and the codomain is the positive real numbers.   So we want to find a positive real number that doesn't equal $3^x$, correct? 
 A: Incorrect. A positive real number that does not equal $3^x$ for any $x$ does not exist.
$f(x) = 3^x$, viewed as a function from $\mathbb R$ to $\mathbb (0,\infty)$ IS onto. Every positive real number can be attained as $3^x$ for some real $x$.
$f(x)=3^x$ as a function from $\mathbb R$ to $\mathbb R$ is NOT onto. You can probably see what real numbers cannot be reached, since you already know that for each $x$, $3^x\in(0,\infty)$.
A: Proof that 
$$f:\Bbb R\to (0,\infty)\;,\;\;\;f(x):=3^x\;,\;\;\;\text{is onto}:$$
$$\forall y\in (0,\infty)\;,\;\;3^x=y\iff x=\log_3y=\frac{\log y}{\log 3}$$
and we're done as we know the (napierian or whatever) logarithm is defined on (all) the positive reals and $\;\log 3\neq 0\;$
A: The question "Prove $f(x)=3^x$ is not onto" is not a complete question. You have to know the sets $A$ and $B$ such that $f: A\rightarrow B$. It is reasonable, though, to expect that the intended question was:
Give a counter-example to prove that
$$f: \mathbb{R}\rightarrow\mathbb{R} $$
$$f(x) = 3^x $$
is not onto. For this it is easy to find a $y\in\mathbb{R}$ such that $f(x)=y$ is impossible for $x\in\mathbb{R}$.
Let $y = -1$ then there exists no $x\in\mathbb{R}$ such that $f(x)=3^x=-1$. This is a counter-example and we are done.
A: If you take $f:\mathbb{R}\to \mathbb{R}$, then $f(x)=3^x$ is not onto for a simple reason. 
Defn : A function $f:A\to B$ is onto if B=ran(f), and 
$ran(f)=\{b\in B:\exists a \in A \ni f(a)=b\}$.  
Now $3^x\ne 0, \forall x\in \mathbb{R}$, which means $\nexists x \in \mathbb{R}$ for which $3^x=0$, so B $\ne ran(f)$
f is thus not onto. $\blacksquare$
