Any reason an exponential decay function approaches but doesn't cross the x-axis?

Question

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?

Answer 1

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 $x$.

Answer 2

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 $c=0$.

Answer 3

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 $x \in \mathbb R$