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)**

Power Expansion entails multiplying out the number base using their base values.

- First, write the place values starting from the right-hand side.
- Write each digit under its place value.
- Multiply each digit by its
**base**raised to the corresponding place value.

(i.e.**base**^{place value},**Note:**for this question, it will be 8^{place value}) - Add up the products. The answer will be the decimal number in base ten.

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} âˆ´ it’s in base 8.

8^{3} | 8^{2} | 8^{1} | 8^{0} |

7 | 6 | 0 | 3 |

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

7603_{eight} = (**7** Ã— 8^{3}) + (**6** Ã— 8^{2}) + (**0** Ã— 8^{1}) + (**3** Ã— 8^{0})

= 3584 + 384 + 0 + 3

= 3,971_{ten}

âˆ´ 7603_{eight} = 3,971_{ten}

b. **Alternate Method – (Successive Multiplication)**

**7** Ã— 8 = 56 + **6** = 62 Ã—Â 8Â = 496 + **0** = 496 Ã— 8 = 3968 + **3** = 3,971_{ten} Â

**Solution** **(ii)** 1101_{two} to base 10

a. **First Method: (Power Expansion)**

2^{3} | 2^{2} | 2^{1} | 2^{0} |

1 | 1 | 0 | 1 |

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

1101_{two} = (1 Ã— 2^{3}) + (1 Ã— 2^{2}) + (0 Ã— 2^{1}) + (1 Ã— 2^{0})

= 8 + 4 + 0 + 1

= 13_{ten}

**Alternate Method**

**1** Ã— 2 = 2 + **1** = 3 Ã— 2 = 6 + **0** = 6 Ã— 2 = 12 + **1** = 13_{ten} Â

**Solution** **(iii)** 5B9_{Twelve} to base 10

a. **First Method: (Power Expansion)**

12^{2} | 12^{1} | 12^{0} |

5 | B | 9 |

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

5B9_{Twelve} = (**5** Ã— 12^{2}) + (**11** Ã— 12^{1}) + (**9** Ã— 12^{0})

= 720 + 132 + 9

= 861_{ten}

**Alternate Method**

**5** Ã— 12 = 60 + **11** = 71 Ã—Â 12Â = 852 + **9** = 861_{ten} Â

### 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 |

In the conversion of a decimal number to base 10, the whole number is treated as the examples above (i.e from the last digit the numbering is 0, 1, 2, 3, etc up to the first digit).

The decimal part on the other hand is numbered -1, -2, -3, etc immediately after the decimal point. The first number after the decimal point will be -1, the second number will be -2, etc.

Hence. We multiply the first number by the base raised to the power minus one (x^{-1}) and so on.

**Solution**

1101.101_{two }= (**1** Ã— 2^{3}) + (**1** Ã— 2^{2}) + (**0** Ã— 2^{1}) + (**1** Ã— 2^{0}) + (**1** Ã— 2^{-1}) + (**0** Ã— 2^{-2}) + (**1** Ã— 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 Ã— 2 = 2Â + 1 = 3 Ã— 2 = 6 + 0 = 6 Ã— 2 = 12 + 1 = 13 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

*Fractional part* = 1 Ã— 2 =Â 2 + 0 = 2 Ã— 2 =Â 4 + 1 = 5Â

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}

âˆ´ 0.101_{two}= 5 Ã— 2^{-3} = 5 Ã— \( \frac {1}{8} \) = 0.625

Summing up the results, then

1101.101_{two }= 13 + 0.625

1101.101_{two }= 13.625_{ten}

### Example 3

Convert **(i)** 4A3D.2_{16} to base 10**(ii)** 4B3.A6_{12} to base 10

**Solution** **(i)** 4A3D.2_{16} to base 10

16^{3} | 16^{2} | 16^{1} | 16^{0} | 16^{-1} | |

4 | A | 3 | D | . | 2 |

Remember D = 13 from our Number Bases table

4A3D.2_{16 }= (**4** Ã— 16^{3}) + (**A** Ã— 16^{2}) + (**3** Ã— 16^{1}) + (**D** Ã— 16^{0}) + (**2** Ã— 16^{-1})

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

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

= 19,005.125

**Solution** **(ii)** 4B3.A6_{12} to base 10

12^{2} | 12^{1} | 12^{0} | 12^{-1} | 12^{-2} | |

4 | B | 3 | . | A | 6 |

A = 10, B = 11

4B3.A6_{12} = (**4** Ã— 12^{2}) + (**B** Ã— 12^{1}) + (**3** Ã— 12^{0}) + (**A** Ã— 12^{-1}) + (**6** Ã— 12^{-2})

= (576) + (11 Ã— 12) + (3) + (11 Ã— 0.0833) + (6 Ã— 0.00694)

= 576 + 132 + 3 + 0.9163 + 0.04164

= 711.96

Super helpful! First method is so easy, I got confused in class when I was learning this.