Any reason an exponential decay function approaches but doesn't cross the x-axis? I've seen graphs of exponential decay functions (where a>0 and 0 is less than b is less than 1) and they don't seem to cross the x-axis.  I think it's true.  Any reason this happens?
 A: You're right, in the sense that if $a \neq 0$ and $b > 0$ are real numbers, then the graph $y = a \cdot b^{x}$ does not cross the $x$-axis. It's enough to show $b^{x} \neq 0$ for all $x$.
The reason comes down to the law of exponents
$$
b^{x + x'} = b^{x} \cdot b^{x'}\quad\text{for all $x$ and $x'$.}
$$
If $b^{x_{0}} = 0$ for some $x_{0}$, then writing $x = x_{0} + (x - x_{0})$ and applying this identity shows $b^{x} = 0$ for all $x$, contradicting the fact that $b^{1} = b > 0$.
To give a direct argument: Setting $x = 1$ and $x' = 0$ gives
$$
b = b^{1} = b^{1 + 0} = b^{1} \cdot b^{0} = b \cdot b^{0}.
$$
Since $b > 0$ by hypothesis, we have $b^{0} = 1$ by dividing both sides by $b$.
Consequently, for all $x$ we have
$$
1 = b^{0} = b^{x + (-x)} = b^{x} \cdot b^{-x}.
$$
Since this equation would be false if $b^{x} = 0$ for some $x$, we deduce that $b^{x} \neq 0$ for all $x$.
A: Mathematically, the answer will never actually be $0$, but it will get very close to $0$. because $f(x)=(c)e^{-kx}$ (where c is a constant) will never be $0$ unless $c=0$.
A: It is known that $b^x=e^{ln(b) \cdot x}$.
And $$e^x=\lim_{n \to \infty} \left (1+\frac{x}{n} \right)^n$$
$1+\frac{x}{n}$ is always positive for $x \in \mathbb R$ 
Thus, $e^{ln(b) \cdot x}$ does not cross the x-axis for $x \in \mathbb R$ 
