1 Section
17 Lessons
4 Weeks

Expand all sectionsCollapse all sections

Exponents – Running Notes
17

Exponents (Level 1)

OffOn

Curriculum

Worked Examples

Example 1

Simplify:

$(2^{20} \div 2^{15}) \times 2^3$

Using quotient law:

$2^{20} \div 2^{15} = 2^{20-15} = 2^5$

Now,

$2^5 \times 2^3 = 2^{5+3} = 2^8$

Therefore,

$(2^{20} \div 2^{15}) \times 2^3 = 2^8$

Example 2

Simplify:

$(3^{-7} \div 3^{-10}) \times 3^{-5}$

Using quotient law:

$3^{-7} \div 3^{-10} = 3^{-7-(-10)} = 3^3$

Now,

$3^3 \times 3^{-5} = 3^{3+(-5)} = 3^{-2}$

Using negative exponent law:

$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

Therefore,

$(3^{-7} \div 3^{-10}) \times 3^{-5} = \frac{1}{9}$

Example 3

Simplify:

$\frac{(2^2)^3 \times 2^5}{2^7}$

First simplify:

$(2^2)^3 = 2^6$

So,

$\frac{(2^2)^3 \times 2^5}{2^7} = \frac{2^6 \times 2^5}{2^7}$

$= \frac{2^{11}}{2^7}$

$= 2^{11-7}$

$= 2^4$

Therefore,

$\frac{(2^2)^3 \times 2^5}{2^7} = 2^4$

Example 4

Simplify:

$\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}$

Write each number in prime factor form:

$12 = 2^2 \times 3$

$9 = 3^2$

$4 = 2^2$

$6 = 2 \times 3$

$8 = 2^3$

$27 = 3^3$

So,

$\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}$

$= \frac{(2^2 \times 3)^4 \times (3^2)^3 \times 2^2}{(3 \times 2)^3 \times (2^3)^2 \times 3^3}$

$= \frac{2^8 \times 3^4 \times 3^6 \times 2^2}{3^3 \times 2^3 \times 2^6 \times 3^3}$

$= \frac{2^{10} \times 3^{10}}{2^9 \times 3^6}$

$= 2^{10-9} \times 3^{10-6}$

$= 2 \times 3^4$

$= 2 \times 81$

$= 162$

Therefore,

$\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27} = 162$

Prev Previous Formulae Summary

Next Worked Examples Part 2 Next

Search

Search

Modal title

Main Content