Division law with variable bases and integer negative powers

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

$a^x\div a^y=a^{x-y}$ax÷​ay=ax−y

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=a2−5 which can be simplified to $a^{-3}$a−3.

 

Examples

Question 1

Simplify the following, giving your answer in index form: $\frac{35x^2y^2}{-5x^{14}y^{14}}$35x2y2−5x14y14​.

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

Question 4