Here is a summary of the index laws we know so far:
The multiplication law: a^m\times a^n=a^{m+n}
The division law: a^m\div a^n=\dfrac{a^m}{a^n}=a^{m-n}, a\neq 0
The power of a power law: \left(a^m\right)^n=a^{m\times n}
The zero power law: a^0=1,a\neq 0
Combining one or more of these laws, together with the order of operations, we can simplify more complicated expressions involving powers. Let's look at some examples to see how.
Fill in the blank to make the equation true.11^{11} \times 11^{⬚}=11^{19}
Simplify \dfrac{\left(17^5\right)^8}{17^{32}}, giving your answer in index form.
Summary of index laws:
The multiplication law: a^m\times a^n=a^{m+n}
The division law: a^m\div a^n=\dfrac{a^m}{a^n}=a^{m-n}, a\neq 0
The power of a power law: \left(a^m\right)^n=a^{m\times n}
The zero power law: a^0=1,a\neq 0
The same rules apply when we are dealing with negative bases, we just need to take care if we are asked to evaluate. We know that the product of two negative numbers is positive, and the product of a positive and a negative number is negative. This means we need to be extra careful when evaluating powers of negative bases.
A negative base raised to an even power will evaluate to a positive answer.
For example \left(-3\right)^4=3^4=81
\left(-3\right)^4\neq -3^4
A negative base raised to a odd power will evaluate to a negative answer.
For example \left(-2\right)^5=-2^5=32
\left(-2\right)^5=-2^5
Using index laws, evaluate \left(-4\right)^{11}\div \left(-4\right)^7.
A negative base raised to an even power will evaluate to a positive answer. (-3)^2=9
A negative base raised to a odd power will evaluate to a negative answer. (-3)^3=-27
When raising a fractional base, we apply the power to both the numerator and the denominator.
Let's consider a simple example like \left(\dfrac{1}{2}\right)^2. This expands to \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1\times 1}{2\times 2}, which evaluates to \dfrac{1}{4} or \dfrac{1^2}{2^2}.
Similarly, a slightly harder expression like \left(\dfrac{2}{3}\right)^3expands to \dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3} giving us \dfrac{2\times 2\times 2}{3\times 3\times 3}. So we can see that \left(\dfrac{2}{3}\right)^3=\dfrac{2^3}{3^3}.
This can be generalised to give us the following rule:
For any base number of the form \dfrac{a}{b}, and any number n as a power, \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}
Simplify \left(\dfrac{23}{41}\right)^8.
For any base number of the form \dfrac{a}{b}, and any number n as a power, \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}