Proof that 2 funtion of the type $f=?iso-8859-1?Q?=5Ft?=(x) = \frac{1}{t} \cdot e^{-tx²} $ don't intersect

Proof that 2 funtion of the type $f_t(x) = \frac{1}{t} \cdot e^{-tx²} $
don't intersect

so as the titles states I wan't to proof that no two functions $f_t$ of
type $$f_t(x) = \frac{1}{t} \cdot e^{-tx²} \; \text{given that} \; t>0$$
share a point. This is a question from my textbook. However I come to the
conclusion that there should be 2 functions should share some point.
$$f_a(x) = \frac{1}{a} \cdot e^{-ax²}$$ $$f_b(x) = \frac{1}{b} \cdot
e^{-bx²}$$ Let $f_a(x) = f_b(x)$
$$\frac{1}{a} \cdot e^{-ax²} = \frac{1}{b} \cdot e^{-bx²}$$ $$\frac{1}{a}
\cdot e^{a} = \frac{1}{b} \cdot e^{b}$$ $$ae^{-a}=be^{b}$$ In order to
proof that the 2 functions don't share a point I would need to proof that
a=b This would mean that the function $g(x)=xe^x$ is a one to one
function. However this is not true!
Now, all continuous one-one functions have to be monotonic (strictly
increasing or decreasing). But we know that this is not the case for
$g(x)$ since $\lim_{x \to 0} g(x)=\lim_{x \to +\infty}f(x)=0$ whereas
$f(1)>0$
Where is my mistake?