Although this calculator says "infinity", you can get a better answer by using logs.
Let x = 72015
Then log(x) = log(72015)
---> log(x) = 2015·log(7)
---> log(x) = 1702.87 (approx)
---> x = 101702.87
Since 10.87 is approximately 7,
the value is a 7 followed by 1702 zeros (approximately).
Although this calculator says "infinity", you can get a better answer by using logs.
Let x = 72015
Then log(x) = log(72015)
---> log(x) = 2015·log(7)
---> log(x) = 1702.87 (approx)
---> x = 101702.87
Since 10.87 is approximately 7,
the value is a 7 followed by 1702 zeros (approximately).