You need to remember how to compose two functions, such that:

`(fog)(x) = f(g(x))`

Notice that the variable x of the function `f(x) ` is replaced by the function `g(x), ` hence, `g(x)` becomes the variable of function `f(g(x)).`

Considering the function `f(x) = x - 1` and the function `g(x) = 2^x` , you may compose the functions `f(x)` and `g(x)` such that:

`(fog)(x) = f(g(x))`

You need to substitute `g(x)` for x in equation of the function `f(x)` such that:

`f(g(x)) = g(x) - 1`

Substituting `2^x` for `g(x)` yields:

`f(g(x)) = 2^x- 1`

`(gof)(x) = g(f(x))`

You need to substitute `f(x)` for x in equation of the function `g(x)` such that:

`g(f(x)) = 2^(f(x))`

Substituting `x - 1` for `f(x) ` yields:

`g(f(x)) = 2^(x-1) => g(f(x)) = (2^x)/2`

Notice that `(fog)(x) != (gof)(x).`

Composing `(fog)(x)` or `(gof)(x)` yields that the variables are either the function `g(x), ` or the function `f(x),` and the elementary functions involved are the linear function `f(x) = x - 1` and the exponential function `g(x) = 2^x` .

**Hence, evaluating the compositions of the functions considered yields `f(g(x)) = 2^x - 1` and `g(f(x)) = (2^x)/2` and `(fog)(x) != (gof)(x).` **