In base 10, the positional of each digit represents which power of 10 that position is taken to and the number represents how many of them you have, i.e. the number 532
100s, 10s, 1s 5.........3......2
Any other base works the same way but instead of powers of 10, we use whatever power that base is, so the number 25 in base 2:
16s, 8s, 4s, 2s, 1s 1.......1......0....0....1
To convert from base 10 to a different base, first find the highest power (position) in that base that is less than your numbers, you can do that by taking the log and round down (for the power) and take ithe base to that number (for the value). E.g. for the number 83, in base 2 you would do log2(83)=6 (rounded down), so that has the 2^6=64s place. Again with base 53, log3(83)=4, and is the 3^4= 81s place. (You can do log(83)/log(2)=6 if you can't change the base on your calculator).
So, starting with the highest you have just found, list all the powers of the base (just divide by the base each time), so base 2:
64s, 32s, 16s, 8s, 4s, 2s, 1s
Next you just go from highest to lowest, and, starting with your number, see how many of each position you can fit into it (keeping the number inside your base), subtracting the amount from it each time you do so. E.g. for base 2, number 83:
In base 3, for the number 140: 81s, 27s, 9s, 3s, 1s 1........2......0....1.....2 140..59......5...2....0
Starting at 140, 1 81 will fit into it so place a 1 in the 81s place and subtract 1*81 from 140 to get 59. 2 27s fit into 59 so write 2 under the 27s place and subtract 2*27=54 from 59 = 5. Continue and you should always end up with 0 remaining.
So 140 in base 3 is 12012.
I hope you understand this, it looks abit complicated at first but it's quite easy when you get the hang of it.
The equation now is 12b + 10 is equal to 2 + 4b take 2 from both sides 12b + 8 is equal to 4b take 12b from both sides 8 is equal to -8b divide by 8 1 is equal to -1b
