Let $A,X,Y$ be topological spaces. Given a function $g: A \times X \to Y$, we can make a corresponding function $\bar g: A \to Y^X$ as $\bar g(a) (x)=g(a,x), \forall a \in A, x \in X$, and vice-versa. The "exponential topology" on $C(X,Y)$ (the set of all continuous functions from $X$ to $Y$) is a topology on it such that for every topological space $A$, $g \in C(A\times X,Y) \iff \bar g \in C(A,C(X,Y))$ .
I can show that if an exponential topology on $C(X,Y)$ exists then it is unique.
My question is: Does there exists an exponential topology on $C(X,Y)$ when $X=Y=[0,1]$ ?
You should consult any book on general topology which covers function spaces, for example
Engelking, Ryszard. "General topology." (1989)
The correspondence $g \mapsto \overline{g}$ is defined for all (not necessarily continuous) functions $g : A \times X \to Y$. It is called the exponential map $\Lambda : \mathcal{F}(A \times X, Y) \to \mathcal{F}(A, \mathcal{F}(X,Y))$. Here, $\mathcal{F}(M,N)$ denotes the set of all functions $M \to N$. If we restrict to continuous functions, we shall consider topologies $\mathfrak{T}_{X,Y}$ on $C(X,Y)$. Let us define
(1) $\mathfrak{T}_{X,Y}$ is proper if $\Lambda(C(A \times X,Y) \subset C(A, (C(X,Y),\mathfrak{T}_{X,Y}))$ for all $A$.
(2) $\mathfrak{T}_{X,Y}$ is admissible if $\Lambda^{-1}(C(A, (C(X,Y),\mathfrak{T}_{X,Y}))) \subset C(A \times X,Y)$ for all $A$.
(3) $\mathfrak{T}_{X,Y}$ is acceptable if it is both proper and admissible (this is what you call exponential topology).
It is a well-known fact that there exist at most one acceptable topology on $C(X,Y)$. Usually function spaces are endowed with the compact-open topology and it is well-known that the compact-open topology is
(a) proper for all spaces $X,Y$,
(b) admissible for all locally compact spaces $X$ and all spaces $Y$.
This answers your question in the affirmative.