Mocuto's Awesome Solution to Sunie's Homework, in O(2^d - 1)) complexity

Sunie.scala

object Sunie {
	def countFrom0to(num : Int, target : Int = 7) : Int = {
		if(num < 10) {
			if (num >= target) 1 else 0
		}
		else {
			val place = logb10(num)
			val count = countFrom0to(num - (num / math.pow(10, logb10(num))).toInt * math.pow(10, logb10(num)).toInt, target)
			val leadDig = num / math.pow(10, place).toInt
			val targetFollowedByZeros = target * math.pow(10, place)
			val multipleTargetPerNum = if (num >= targetFollowedByZeros) math.min(math.pow(10, place), num - targetFollowedByZeros + 1).toInt else 0
			count + (leadDig * countFrom0to(math.pow(10, place).toInt - 1, target)).toInt + multipleTargetPerNum
		}
	}

	def countTargetInRange(a : Int, b : Int, target : Int = 7) : Int = { // Where b > a
		val aCount = countFrom0to(a, target)
		val bCount = countFrom0to(b, target)
		bCount - aCount
	}

	def logb10(x : Int) : Int = (math.log(x) / math.log(10)).toInt

	def lazyVersion(a : Int, b : Int, target : Int = 7) : Int = { //where b > a
		(0 to b).map(_.toString).mkString.filter(_ == (target.toString())(0)).length - (0 to a).map(_.toString).mkString.filter(_ == (target.toString())(0)).length
	}

	def main(args : Array[String]): Unit = {
		assert(lazyVersion(0, 999) == countTargetInRange(0, 999))
		assert(lazyVersion(100, 150) == countTargetInRange(100, 150))
		assert(lazyVersion(10, 777) == countTargetInRange(10, 777))
	}

}

Iunno scala but why (0 to b)... - (0 to a)...? can't you go (a to b)...? on line 25
EDIT: oh you do that for the good version too. maybe i don't get the problem