Considering the fact that `f(2) = 1` , it means `x = 2`

and the function is equal to `1` , we will have:

`a(2)^r = 1` (equation 1)

For the given `f(32) = 4` , we will have:

`a(32)^r = 4 ` (equation 2)

Solving for `r` using the first equation:

`a(2)^r = 1 `

Divide both sides by `2^r`

`(a(2)^r)/((2)^2) = 1 /(2)^r`

Substituting the `1/(2)^2` on the second equation:

`a(32)^r = 4 `

`(1/((2)^r))*(32)^r = 4`

`(32)^r/(2)^r = 4 `

Applying the identity: `m^x/n^x=((m)/(n))^x` .

`(32)^r/(2)^r = 4`

`((32)/(2))^r = 4`

Applying division:

`(16)^r = 4`

We can rewrite the `16` as `4^2` .

`((4)^2)^r = 4`

Applying the identity `(m^x)^y = m^(x*y)` .

`4^(2r) = 4`

We can write the `4` as `4^1` .

`4^(2r) = 4^1`

Since we have the same bases, we will have:

`2r = 1`

Divide both sides by `2` .

Therefore,` r = 1/2` .