New Zealand
Level 7 - NCEA Level 2

# Division law with variable bases and negative powers

Lesson

We've already learnt about the division law which states:

$a^x\div a^y=a^{x-y}$ax÷​ay=axy

But what happens when the $y$y value is larger than the $x$x value? Well we subtract them in just the same way except we'll end up with a negative answer. For example, $a^2\div a^5=a^{2-5}$a2÷​a5=a25 which can be simplified to $a^{-3}$a3.

#### Examples

##### Question 1

Simplify the following, giving your answer in index form: $\frac{35x^2y^2}{-5x^{14}y^{14}}$35x2y25x14y14.

Think: Let's separate the terms and apply the Index division law.

Do

 $35\div\left(-5\right)$35÷​(−5) $=$= $-7$−7 the coefficient term $x^2\div x^{14}$x2÷​x14 $=$= $x^{2-14}$x2−14 the $x$x's $=$= $x^{-12}$x−12 $y^2\div y^{14}$y2÷​y14 $=$= $y^{2-14}$y2−14 the $y$y's $=$= $y^{-12}$y−12 $\frac{35x^2y^2}{-5x^{14}y^{14}}$35x2y2−5x14y14​ $=$= $-7x^{-12}y^{-12}$−7x−12y−12 all together

##### Question 2

Convert the following to a fraction and simplify using the index laws: $\left(-5x^6\right)\div\left(-3x^8\right)\div\left(-5x^6\right)$(5x6)÷​(3x8)÷​(5x6).

Think: Dividing by a fraction is the same as multiplying by its reciprocal. We will then solve this problem in a couple of steps

Do:

 $\left(-5x^6\right)\div\left(-3x^8\right)\div\left(-5x^6\right)$(−5x6)÷​(−3x8)÷​(−5x6) $=$= $\frac{-5x^6}{-3x^8}\times\frac{1}{-5x^6}$−5x6−3x8​×1−5x6​ $=$= $\frac{-5x^6}{15x^{14}}$−5x615x14​ $=$= $\frac{-1}{3x^8}$−13x8​

##### Question 3

Convert the following to a fraction and simplify using the index laws:

$\left(-240u^{32}\right)\div\left(-8u^9\right)\div\left(-5u^{12}\right)$(240u32)÷​(8u9)÷​(5u12)

##### Question 4

Simplify the following, giving your answer with a positive index: $\frac{9x^3}{3x^{-4}}$9x33x4

### Outcomes

#### M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

#### 91261

Apply algebraic methods in solving problems