(car (cons x y))

(car (lambda (m) (m x y)))

((lambda (m) (m x y)) (lambda (p q) p)))

(lambda (x y) x)))

x

(define (cdr z) (z (lambda (x y) y)))



2.5 - Given that we know the bases - 2, and 3, we only need to find the exponents.

For example, 6 can only be represented one way - 2^1 3^1, meaning our pair of numbers a and b are 1

So, cons is simple:

(define (cons x y) (* (expt 2 x) (expt 3 y)))

Now we're left solving for a and b

log(2^a * 3^b) = alog(2) + blog(3)

call our "cons" value x

Because integer factorizations are unique we only need to find the biggest power of two and three that 
go into the number.

(define (power-of-x value base) 
  (define (iter current-power)
    (if (< ((expt base current-power) value)
	(iter (+ current-power 1))
	(current-power))))

Then with this we can write car and cdr as

(define (car int) (power-of-x int 2))
(define (cdr int) (power-of-x int 3))