Converting from other Bases to Base 10

Here, we will learn the methods to convert from other bases to Base 10

Example 1

Convert

(i) 7603_eight to base 10

(ii) 1101_two to base 10

(iii) 5B9_twelve to base 10

Solution (i) 7603_eight to base 10

a. First Method: (Power Expansion)

Students often find it difficult to locate the place values. The easiest way is to start from the last digit to the first digit. The last Digit always starts with a 0, the digit that follows a 1, and so on. (0, 1 2, 3, etc)

3	2	1	0
7	6	0	3

Write down the base raised to the place value. The number we are converting is 7603_eight ∴ it’s in base 8.

83	82	81	80
7	6	0	3

Multiply each digit by its base raised to the corresponding place value.

7603_eight = (7 × 8³) + (6 × 8²) + (0 × 8¹) + (3 × 8⁰)

              = 3584 + 384 + 0 + 3

              = 3,971_ten

               ∴    7603_eight = 3,971_ten

b. Alternate Method – (Successive Multiplication)

7 × 8 = 56 + 6 = 62 × 8  = 496 + 0 = 496 × 8 = 3968 + 3 = 3,971_ten  

Solution (ii) 1101_two to base 10

a. First Method: (Power Expansion)

23	22	21	20
1	1	0	1

Multiply each digit by its base raised to the corresponding place value

1101_two = (1 × 2³) + (1 × 2²) + (0 × 2¹) + (1 × 2⁰)

= 8 + 4 + 0 + 1

= 13_ten  

Alternate Method

1 × 2 = 2 + 1 = 3 × 2 = 6 + 0 = 6 × 2 = 12 + 1 = 13_ten  

Solution (iii) 5B9_Twelve to base 10

a. First Method: (Power Expansion)

122	121	120
5	B	9

Multiply each digit by its base raised to the corresponding place value (Note: B =11)

5B9_Twelve = (5 × 12²) + (11 × 12¹) + (9 × 12⁰)

= 720 + 132 + 9

= 861_ten  

Alternate Method

5 × 12 = 60 + 11 = 71 × 12  = 852 + 9 = 861_ten  

Example 2

Convert 1101.101_two to base ten

\( \scriptsize 2^{3} \)	\( \scriptsize 2^{2} \)	\( \scriptsize 2^{1} \)	\( \scriptsize 2^{0} \)		\( \scriptsize 2^{-1} \)	\( \scriptsize 2^{-2} \)	\( \scriptsize 2^{-3} \)
1	1	0	1	.	1	0	1

Solution

1101.101_two = (1 × 2³) + (1 × 2²) + (0 × 2¹) + (1 × 2⁰) + (1 × 2^-1) + (0 × 2^-2) + (1 × 2^-3)

= 8 + 4 + 0 + 1 + \( \frac {1}{2} \) + 0 + \( \frac {1}{8} \)

= 8 + 4 + 0 + 1 + 0.5 + 0 + 0.125

=  13.625

1101.101_two= 13.625_ten

Alternate Method

Whole number part = 1 × 2 = 2 + 1 = 3 × 2 = 6 + 0 = 6 × 2 = 12 + 1 = 13                                      

Fractional part = 1 × 2 = 2 + 0 = 2 × 2 = 4 + 1 = 5 

The result for the whole number part is: 1101_two = 13_ten

The place value of the last column of the fractional part is 2^-3

  ∴   0.101_two= 5 × 2^-3 = 5 × \( \frac {1}{8} \) = 0.625

Summing up the results, then

1101.101_two = 13 + 0.625

1101.101_two = 13.625_ten

Example 3

Convert

(i) 4A3D.2₁₆ to base 10

(ii) 4B3.A6₁₂ to base 10

Solution (i) 4A3D.2₁₆ to base 10

163	162	161	160		16-1
4	A	3	D	.	2

Remember D = 13 from our Number Bases table

4A3D.2₁₆ = (4 × 16³) + (A × 16²) + (3 × 16¹) + (D × 16⁰) + (2 × 16^-1)

= (16,384) + (10 × 256) + (48) + (13 × 1) + (2 × 0.0625)

= 16,384 + 2,560 + 48 + 13 + 0.125

= 19,005.125

Solution (ii) 4B3.A6₁₂ to base 10

122	121	120		12-1	12-2
4	B	3	.	A	6

A = 10, B = 11

4B3.A6₁₂ = (4 × 12²) + (B × 12¹) + (3 × 12⁰) + (A × 12^-1) + (6 × 12^-2)

= (576) + (11 × 12) + (3) + (10 × 0.0833) + (6 × 0.00694)

= 576 + 132 + 3 + 0.833 + 0.04164

= 711.875