Decision problems (checking whether an instance of a problem has a solution or not) and Search problems (actually finding a solution of an instance of a problem if it has one) are two sides of the same coin.
Suppose that you have a computable function $f : \mathbb{N} \to \mathbb{N}$.
Then you can easily convert it to a decision version using this language:
$L = \{ (x,n) \mid f(x) \leq n \}$
Clearly if you can solve the decision problem $(x,n) \in^? L$ you can compute $f(x)$ in $O(\log f(x))$ steps using a binary search.
Usually the equivalent decision problem is formulated in a manner that reflects the structure of the source problem; a well-known example is boolean satisfiability.
"Given a boolean formula $\varphi$ in $n$ variables $x_1, x_2, ..., x_n$ calculate a satisfying assignment if it exists"
This can be solved using the decision version: "Given a boolean formula $\varphi$, is it satisfiable?" (i.e. $\varphi \in^? SAT$)
You can start setting $x_1 = True$ and checking if $\varphi_{x_1=T}$ (the formula $\varphi$ in which all $x_1$ occurrences have been replace with $True$ is satisfiable). If the answer is $Yes$ ($\varphi_{x_1=T} \in SAT$) then set $x_1 = True$, $x_1 = False$ otherwise.
Then continue with $x_2, x_3, ...$ up to $x_n$ and at the end you'll have a satisfying assignment, or the procedure will stop at some $x_i$ for which $\varphi_{...,x_i=T} \notin SAT$ and $\varphi_{...,x_i=F} \notin SAT$ (i.e. the formula is unsatisfiable).