A given number can be converted from any base to base 10 using the **Expansion Method**.

Once you are able to work out the** place values**, you can easily convert other base numbers to binary.

Let’s take a look at the following examples;

### Example 1

Write the following numbers in expanded form. **1.** 356_{eight}**2.** 10321_{four }

**Solution **

**1st Step: **To use the expansion method for conversion we first find the place numbers.

**1.** 356_{eight }

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)

**2nd Step:** Multiply each digit by its base number raised to the power of its place value

356_{eight } = 3 x 8^{2} + 5 x 8^{1} + 6 x 8^{0}

Base eight is also called the octal system

**2.** 10321_{four}

Multiply each digit by its base number raised to the power of its place value

10321_{four} = 1 x 4^{4} + 0 x 4^{3} + 3 x 4^{2} + 2 x 4^{1} + 1 x 4^{0}

To convert any number base to base ten, simply express it as the power of the given base. As discussed earlier the place value is important here.

### Example 2

Express the following binary numbers as numbers in base ten.** ****a.** 110100_{two}**b. **534_{six}**c. **4307_{eight}

**a.** 110100_{two}

**Solution **

Using the expanding method

Place Value | ^{5} | ^{4} | ^{3} | ^{2} | ^{1} | ^{0} |

Binary Number | 1 | 1 | 0 | 1 | 0 | 0 |

110100_{ten} = 1 x 2^{5} + 1 x 2^{4} + 0 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 0 x 2^{0}

= (1 x 2^{5}) + (1 x 2^{4}) + (0 x 2^{3}) + (1 x 2^{2}) + (0 x 2^{1}) + (0 x 2^{0})

= (1 x 32) + (1 x 16) + 0 + (1 x 4) + 0 + 0

= 32 + 16 + 0 + 4 + 0 + 0

= 32 + 16 + 4

= 52_{ten}

110100_{two} = 52_{ten }

**b.** 534_{six}

**Solution**

Using the expansion method:

Place Value | ^{2} | ^{1} | ^{0} |

Binary Number | 5 | 3 | 4 |

534_{six} = 5 x 6^{2} + 3 x 6^{1} + 4 x 6^{0}

= (5 x 36) + (3 x 6) + (4 x 1)

= 180 + 18 + 4

= 202_{ten}

534_{six} = 202_{ten }

**Note**

6^{2 }= 6 x 6 = 36

6^{1 } = 6

6^{0} = 1

**c.** 4307_{eight}

**Solution**

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

Binary Number | 4 | 3 | 0 | 7 |

Using the expanding method: (same approach as Examples above)

4307_{eight} = 4 x 8^{3} + 3 x 8^{2} + 0 x 8^{1} + 7 x 8^{0}

= (4 x 512) + (3 x 64) + (0 x 8) + (7 x 1)

= 2048 + 192 + 0 + 7

= 2247_{ten }

â‡’ 4307_{eight} = 2247_{ten }

### Example 3

Find the value of (111_{two})^{2} in base two

**Solution**

**1st Step: **The first thing to do is to convert the number to base ten.**2nd Step: **After converting to base ten we will then square the value.**3rd Step: ** Finally we will convert the squared value back to base two.

**1st Step:**

Place Value | ^{2} | ^{1} | ^{0} |

Binary Number | 1 | 1 | 1 |

Using the expansion method;

111_{two} = (1 x 2^{2}) + (1 x 2^{1}) + (1 x 2^{0})

= 1 x 4 + 1 x 2 + 1 x 1

= 4 + 2 + 1

= 7_{ten}Â

**2nd Step:**

Recall that (111_{two})^{2 }= (7_{ten})^{2 } = 7 x 7 = 49

= 49_{ten}Â Â

**3rd Step:**

Then convert 49_{ten} to base two (binary) using the method of repeated division

â‡’ 49_{ten }= 110001_{two}

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