diesieben07
11/29/2018, 11:14 PMa * pow(10, log10(b)) + b
Hamza
11/30/2018, 3:23 AMbdawg.io
11/30/2018, 3:37 AMHamza
12/01/2018, 3:38 PMdiesieben07
12/03/2018, 11:01 AMlog10(5) = 0
, log10(140) = 2
, so basically "how many powers of 10 are in there".
So for 5 and 15 you have to multiply 5 by however many lots of 10 are in 15 (1) and then add 15. However I notice now that I am missing a +1 in there, so it should be
a * pow(10, log10(b) + 1) + b
For 5 and 15 this becomes:
5 * pow(10, log10(15) + 1) + 15
which is 5 * pow(10, 1 + 1) + 15
which is 5 * 100 + 15
which is 515
=> the concatenated number.