change of base?

Question

im trying to get my head around change of base
eg73base10, change to base 2
and
1234base5, change to base 10

HELP!!

Answer 1

Write 73 down as a sum of various powers of 2:

73 = 64 + 8 + 1

Say to yourself "that is one 64 and no 32s and no 16s and one 8 and no 4s and no 2s and one 1" and then you can write down what you just said from left to right:

1001001

To go in the other direction you look at this 7 digit binary number and say to yourself (work from right to left) "the denary nu,mber contains a 1 and an 8 and a 64, so that is 73"

The powers of 5 we will need in base-5 maths for a 4-digit number are 1 (units), 5 (in the "tens place"), 25 (in the "hundreds place") and 125 ("in the thousands place"). The largest such number, 4444, would be 624 (denary) as 625 would be 10000,

So 1234 (base-5) is (1 x 125) + (2 x 25) + (3 x 5) + (4 x 1) = 194 (denary). To go in the other direction and answer "what is 194 in base 5?", you need to do some division and remainders:

a) notice 194 is smaller than 625 but larger than 125

b) ask "how many 125s are there in 194?" and write down the answer, working from left to right (1 "in the thousands place")

c) Subtract 125 from 194 to leave a remainder of 69

d) ask "how many 25s are there in 69?" and write down the answer, working from left to right (2 "in the hundreds place")

e) Subtract 50 from 69 to leave a remainder of 19

f) ask "how many 5s are there in 19?" and write down the answer, working from left to right (3 "in the tens place")

g) Subtract 15 from 19 to leave a remainder of 4

h) ask "how many 1s are there in 4?" and write down the answer, working from left to right (4 "in the units place")

and thus 1234 is the answer.

Answer 2

Changing to base 10 is usually easier. If a number is in another base, then each digit is in "place value" starting on the right with that base to the zero power (which equals one), then the next digit to the right is multiplied by that base to the 1st power, then the next is multiplied by the base to the second power, etc...

So for the second question, 1 gets multiplied by five to the third power, 2 gets multiplied by five to the second, etc...

Changing from base 10 to another base is harder - especially to write here online. I found a couple of websites pretty quickly which demonstrate how to do this. The first one looks like the best:

http://courses.cs.vt.edu/~csonline/NumberSystems/Lessons/DecimalToBinaryConversion/index.html

http://www.is.wayne.edu/olmt/binary/page3.htm

I hope this helps! Good luck

Answer 3

Question 1
64--32--16--8--4--2--1
1----0----0---1--0--0--1
from above, 73 base ten = 1001001 base two

Question 2
125--25--5--1
1-----2----3--4

125 + 50 + 15 + 4 = 194 in base ten
1234 base 5 = 194 base ten

128--64--32--16--8--4--2--1
--1----1----0----0--0--0--1--0

1234 base 5 = 194 base ten = 11000010 base two

Answer 4

log[base10] 73 = (log[base2] 73) / (log[base2] 10)