**Exponentiation** is a mathematical operation involving two numbers, **the base $x$** and **the exponent $a$**. When $a$ is a positive integer, exponentiation corresponds to **repeated multiplication of the base**.

For example, if the base is $x=2$ and the exponent $a=3$, it is writen as $x^a=2^3$. This means that we have multiplied the number $2$ by itself three times or: $2\cdot{2}\cdot{2}$, which equals $8$.

By definition, every number which has 0 as its exponent is equal to 1. This means that, no matter how large is the base, if their exponent is equal to 0, that number is always equal to 1.

Every number that doesn’t have an exponent attached to it, actually has the number 1 as its exponent. The number 1 is the default exponent of every number, so it isn’t necessary to write it down, but in some tasks it can be helpful to do so.

One multiplied by one is always one, no matter how many times you repeat the multiplication, so 1 to any power is always equal to 1.

## Negative exponents

If the exponent is a positive integer, exponentiation corresponds to repeated multiplication of the base, so what does it mean if the exponent is a negative integer? The reciprocal value of the base is than used to turn the negative exponent into a positive.

**$a^{-n}=(a^{-1})^n=\left(\frac{1}{a}\right)^n=\frac{1}{a^n}$**

The same goes the other way around. If an unknown is in the denominator, the denominator can become a numerator by changing the sign of the exponent. In some cases, this will prove to be a very useful feature, especially when working with inverse numbers and functions.

* Example 1: *Write these expressions using only positive exponents:

a) $a^{-7}$

b) $\frac{-6x^{-1}}{y^5}$

c) $\frac{-12x^{-6}y^{-9}}{z^3}$

**Solution:**

a) $a^{-7}=\frac{1}{a^7}$

b) $\frac{-6x^{-1}}{y^5}=\frac{-6}{x^1 \cdot y^5}=\frac{-6}{xy^5}$

c) $\frac{-12x^{-6}y^{-9}}{z^3} = \frac{-12}{x^6 \cdot y^9 \cdot z^3}$

## Addition

How does one add or subtract exponents?

For example, $\ 2^2 = 4$ and $\ 2^3 = 8$ so $\ 4 + 8 = 12$. But for $\ 2^2 + 2^3$, the answer is not that obvious. One cannot add nor subtract numbers that have different exponents or different bases.

Most interesting tasks involve unkowns, but the same rules apply to them.

Let’s look at a simple equation:

Since $\ x = x^1$ and $\ 1 = x^0$ we can write our equation like this:

How would you normally solve it? The variables with $x$ are added separately, and separately variables without $x$.

The same will apply to larger exponents:

$\ x^{12} + 3 \cdot {x^{12}} + 2 \cdot {x^2} = 4 \cdot {x^{12}} + 2 \cdot {x^2}$

* Example 2: *Add exponents

## Subtraction

The same rules that apply to adding exponents, apply to subtracting as well.

You can only subtract numbers that have unknowns with the same exponent.

* Example 3:* Subtract exponents:

$ 4x^{12} – 0.25 x^4 + 2x^2 – 3x^2 – 3x^{12} = ?$

**Solution:**

$ (4x^{12} – 3x^{12}) – 0.25\cdot {x^4} + (2x^2 – 3x^2) = x^{12} – 0.25\cdot {x^4} – x^2$

## Multiplication

There are two basic rules for multiplication of exponents.

The first rule – **if bases are the same, their exponents are added together**.

For example: $\ 2^2 \cdot {2^3} = 2^{2 + 3} = 2^5$.

Similarly, with a **negative exponent**, it can either be left as it is, or transformed into a **reciprocal fraction**.

For example: $\ 2^{-2} \cdot {2^{-3}} = 2^{- 2 – 3} = 2^{-5} = \left(\frac{1}{2}\right)^5$.

The second rule – if bases are different, but exponents are the same, bases are multiplied and exponents remain the same.

For example: $\ 2^2 \cdot {3^2} = (2 \cdot {3})^2 = 6^2$.

**Example 4:**

$ 2^2 \cdot {4^2} = ?$

**Solution:**

To multiply two exponents, their base or their exponents must be the same. In this example, neither is the case. So, the first step is to, whenever possible, to turn every number to the lowest base. In this example the number $4$ can be written as $2^2$.

$ 2^2 \cdot {(2^2)^2} = ?$

The square represents the number multiplied by itself so $\ (2^2)^2$ can be written as $\ 2^2 \cdot {2^2} = 2^{2 + 2} = 2^4$.

From Example 4, this generalisation can be made:

Final solution: $\ 2^2 \cdot {4^2}= 2^2 \cdot {(2^2)^2} = 2^2 \cdot {2^4} = 2^{2+4} = 2^6$.

**Example 5:**

$ \left(\frac{2}{3}\right)^2 \cdot {0.2^2} = ?$

**Solution:**

$$=\left (\frac{2}{3}\right)^2 \cdot \left (\frac{2}{10}\right)^2$$

$$=\left (\frac{2}{3}\right)^2 \cdot\left (\frac{1}{5}\right)^2$$

$$= \left(\frac{2}{3} \cdot \frac{1}{5}\right)^2 $$

$$= \left(\frac{2}{15}\right)^2$$

**Example 6:**

$\ (x^2 y^3)(x^5 y^4 )$

*Solution:*

Multiplication is associative so the order of brackets doesn’t make a difference. The factors with the same bases are multiplied as explained before, so their exponents are added.

$ (x^2 \cdot y^3)(x^5 \cdot y^4) = x^2 \cdot x^5 \cdot y^3 \cdot y^4 = x^7 \cdot y^7 = (xy)^7$

## Division

As for multiplication, there are two basic rules for dividing exponents.

The first rule – **when bases are the same, their exponents are subtracted**.

For example: $\ 2^2 : 2 = \frac{2^2}{2} = 2^{2 – 1} = 2^1 = 2$, which can easily be checked since $4 : 2 = 2$.

For example: $\ 2^{-2} : 2^{-1} =\frac{2^{-2}}{2^{-1} }= 2^{-2-(-1)} = 2^{-1} = \frac{1}{2}$.

The second rule – **if bases are different, but exponents are the same, bases are divided and the exponents remain the same**.

For example: $\ 2^2 : 3^2 = \frac{2^2}{3^2 } = (2 : 3)^2 = \left(\frac{2}{3}\right)^2$.

**Example 7:**

$\frac{4^2}{4^3} + \frac{1}{2} = ?$

*Solution:*

$\frac{4^2}{4^3} + \frac{1}{2} = 4^{2 – 3} + \frac{1}{2} = 4^{-1} + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{1 + 2}{4} = \frac{3}{4}$

**Example 8:**

$\frac{4^5}{4^{-2}} – 0.2 \cdot {4} + \frac{1}{2} \cdot {2^8} = ?$

**Solution:**

$\frac{4^5}{4^{-2}} – 0.2 \cdot 4 + \frac{1}{2} \cdot 2^8 = 4^{5 – (-2)} – \frac{2}{10} \cdot 4 + \frac{2^8}{2^1} = 4^{5 + 2} – \frac{1}{5} \cdot 4 + 2^{8 – 1} = 4^7 – \frac{4}{5} + 2^7$

**Example 9:**

$\frac{18x^5y^6a^2}{6xy^2a^5} = ?$

**Solution:**

$\frac{18x^5y^6a^2}{6xy^2a^5} = 3x^{5 – 1}y^{6 – 2}a^{2 – 5} = 3x^4y^4a^{-3} = \frac{3x^4y^4}{a^3}$

If, like in this example, a task involves only division and multiplication, the fraction can be divided into two smaller fractions.

$\frac{x^2y^3 + x^5y}{xy} = \frac{x^2y^3}{xy} + \frac{x^5y}{xy} = xy^2 + x^4$

## Exponents worksheets

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