Division law with variable bases and 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
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}}$9x33x−4