1.05 Index laws
An expression of the form $b^n$bn is called an exponential or power expression. This expression is read "$b$b raised to the $n^{th}$nth power". $b$b is the base and $n$n is the index or exponent.
- When multiplying powers with the same base, add the indices.
$a^m\times a^n=a^{m+n}$am×an=am+n
- When dividing powers with the same base, subtract the indices.
$\frac{a^m}{a^n}=a^{m-n}$aman=am−n
- When raising a power to a power, multiply the indices.
$\left(a^m\right)^n=a^{m\times n}$(am)n=am×n
- When raising a product to a power, we can rewrite this as a product of powers.
$\left(ab\right)^m=a^mb^m$(ab)m=ambm
- When raising a fraction to a power, we can rewrite this as the quotient of the powers.
$\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}$(ab)m=ambm
- Any non-zero number raised to the power of zero is equal to 1.
$a^0=1,\text{for }a\ne0$a0=1,for a≠0
- Raising a non-zero number to a negative power results in the reciprocal of the number raised to a positive power.
$a^{-n}=\frac{1}{a^n}\text{for }a\ne0$a−n=1anfor a≠0
Let's look at some examples that illustrate how these identities are consistent and how to apply them.
Multiplying powers
Consider the expression $a^5a^3$a5a3. Notice that the terms share like bases. Let's think about what this would look like if we expanded the expression:
We can see that there are eight $a$a's being multiplied together, and notice that $8$8 is the sum of the powers in the original expression.
Worked example
Example 1
Simplify the expression $3a^{2x}\times6a^{x+1}$3a2x×6ax+1.
Think: Rearrange to bring the numerical factors to the front followed be the two power expressions. Notice the power expressions have the same base and are multiplied, hence, add the indices.
Do:
| $3a^{2x}\times6a^{x+1}$3a2x×6ax+1 | $=$= | $3\times6\times a^{2x}\times a^{x+1}$3×6×a2x×ax+1 |
| $=$= | $18\times a^{2x+x+1}$18×a2x+x+1 | |
| $=$= | $18a^{3x+1}$18a3x+1 |
Dividing powers
If we wanted to simplify the expression $a^6\div a^2$a6÷a2, we could write it in expanded form as:
Dividing top and bottom by $a$a twice, we are left with $a^4$a4 which is the difference between the two indices in the original expression.
Worked example
Example 2
Simplify the expression $\frac{5a^{3x}\times a^{4x}}{10a^{2x+1}}$5a3x×a4x10a2x+1.
Think: Simplify the top using our multiplication law. Divide the common factor of 5 from top and bottom. Lastly, simplify the powers of $a$a by subtraction, be careful to apply the subtraction to both terms of the index from the denominator.
Do:
| $\frac{5a^{3x}\times a^{4x}}{10a^{2x+1}}$5a3x×a4x10a2x+1 | $=$= | $\frac{5a^{3x+4x}}{10a^{2x+1}}$5a3x+4x10a2x+1 |
| $=$= | $\frac{1a^{7x}}{2a^{2x+1}}$1a7x2a2x+1 | |
| $=$= | $\frac{a^{7x-\left(2x+1\right)}}{2}$a7x−(2x+1)2 | |
| $=$= | $\frac{a^{5x-1}}{2}$a5x−12 |
Raising a power to a power
Consider the expression $\left(5^2\right)^3$(52)3. What is the resulting power of base $5$5? To find out, have a look at the expanded form of the expression:
| $\left(5^2\right)^3$(52)3 | $=$= | $\left(5\times5\right)^3$(5×5)3 |
| $=$= | $\left(5\times5\right)\times\left(5\times5\right)\times\left(5\times5\right)$(5×5)×(5×5)×(5×5) | |
| $=$= | $5^6$56 |
In the expanded form, we can see that $5$5 is multiplied by itself $6$6 times, which was the $5^2$52 multiplied by itself $3$3 times. So we got the total index as $3\times2$3×2.
Product to a power
If we wanted to simplify the expression $\left(ab\right)^4$(ab)4, we could write it in expanded form as:
| $\left(ab\right)^4$(ab)4 | $=$= | $\left(ab\right)\times\left(ab\right)\times\left(ab\right)\times\left(ab\right)$(ab)×(ab)×(ab)×(ab) |
| $=$= | $\left(a\times a\times a\times a\right)\times\left(b\times b\times b\times b\right)$(a×a×a×a)×(b×b×b×b) | |
| $=$= | $a^4b^4$a4b4 |
We can see once we expand we can group the terms with the same base and the result is the same as applying the power to each term in the product.
Practice question
Question 1
Simplify the following, giving your answer in index form:
$\frac{\left(n^8r^5\right)^5}{\left(n^4r\right)^5}$(n8r5)5(n4r)5
Fraction to a power
If we wanted to simplify the expression $\left(\frac{2}{x}\right)^3$(2x)3, we could write it in expanded form as:
| $\left(\frac{2}{x}\right)^3$(2x)3 | $=$= | $\frac{2}{x}\times\frac{2}{x}\times\frac{2}{x}$2x×2x×2x |
| $=$= | $\frac{2\times2\times2}{x\times x\times x}$2×2×2x×x×x | |
| $=$= | $\frac{2^3}{x^3}$23x3 |
We can see once we expand we can group the terms with the same base and the result is the same as applying the power to the numerator and denominator of the fraction.
Power of zero
As we saw when dividing powers with like bases, we subtract the indices.
So what happens when we subtract and we are left with a power of 0? For example,
| $\frac{4^1}{4^1}$4141 | $=$= | $4^{1-1}$41−1 |
| $=$= | $4^0$40 |
However, we can also see the left hand side simplifies to $1$1. Notice that this will also be the case with $\frac{4^2}{4^2}$4242 or any expression where we are dividing like bases whose indices are the same. So the identity of $a^0=1$a0=1 for any non-zero $a$a seems to be consistent with our previous rules.
Reflect: What is $0^0$00? Is there consensus for how it should be defined?
Practice question
Question 2
Simplify $9\times\left(15x^6\right)^0$9×(15x6)0.
Negative powers
Again as we saw when dividing powers with like bases, we can subtract the indices. That is:
$a^x\div a^y=a^{x-y}$ax÷ay=ax−y
But what happens when $y$y is bigger than $x$x? For example, if I simplified $a^3\div a^5$a3÷a5 using the division law, I would get $a^{-2}$a−2. So what does a negative index mean? Let's expand the example to find out:
Remember that when we are simplifying fractions, we are looking to divide out common factors in the numerator and denominator. Remember that any number divided by itself is $1$1.
So using the second approach, we can also express $a^3\div a^5$a3÷a5 with a positive index as $\frac{1}{a^2}$1a2.
Negative powers and fractions
We can combine our rules to help simplify fractions involving negative powers.
Worked examples
Example 3
Write $\left(\frac{2}{5}\right)^{-3}$(25)−3 as a fraction in simplest form.
We can rewrite $\left(\frac{2}{5}\right)^{-3}$(25)−3 as $\left(\left(\frac{2}{5}\right)^{-1}\right)^3$((25)−1)3, then we can take the inverse of the fraction (flip it) before applying the power of three.
Hence,
| $\left(\frac{2}{5}\right)^{-3}$(25)−3 | $=$= | $\left(\left(\frac{2}{5}\right)^{-1}\right)^3$((25)−1)3 |
| $=$= | $\left(\frac{5}{2}\right)^3$(52)3 | |
| $=$= | $\frac{5^3}{2^3}$5323 | |
| $=$= | $\frac{125}{8}$1258 |
Example 4
Simplify $\frac{3a^4b}{6a^7b^{-3}c^{-4}}$3a4b6a7b−3c−4, giving the answer with positive indices.
First using index rules for division combine any terms with common bases and simplify the numerical terms. Look out for negative signs when subtracting.
| $\frac{3a^4b}{6a^7b^{-3}c^{-4}}$3a4b6a7b−3c−4 | $=$= | $\frac{1a^4b}{2a^7b^{-3}c^{-4}}$1a4b2a7b−3c−4 |
| $=$= | $\frac{a^{\left(4-7\right)}b^{\left(1+3\right)}}{2c^{-4}}$a(4−7)b(1+3)2c−4 | |
| $=$= | $\frac{a^{-3}b^4}{2c^{-4}}$a−3b42c−4 |
Lastly to write this without negative fractions, we could write these as a product of the components:
| $\frac{a^{-3}b^4}{2c^{-4}}$a−3b42c−4 | $=$= | $\frac{1}{a^3}\times b^4\div\frac{2}{c^4}$1a3×b4÷2c4 |
| $=$= | $\frac{1}{a^3}\times b^4\times\frac{c^4}{2}$1a3×b4×c42 | |
| $=$= | $\frac{b^4c^4}{2a^3}$b4c42a3 |
Notice that the negative powers swapped from the numerator to the denominator or vice versa. Can you always simply swap them? Another way to see why this happens is by approaching this slightly differently at the start and subtracting the lower powers from the higher powers to ensure you have positive powers, as follows:
| $\frac{3a^4b}{6a^7b^{-3}c^{-4}}$3a4b6a7b−3c−4 | $=$= | $\frac{1a^4b^1c^0}{2a^7b^{-3}c^{-4}}$1a4b1c02a7b−3c−4 |
| $=$= | $\frac{b^{\left(1+3\right)}c^{\left(0+4\right)}}{2a^{\left(7-4\right)}}$b(1+3)c(0+4)2a(7−4) | |
| $=$= | $\frac{b^4c^4}{2a^3}$b4c42a3 |
Practice questions
Question 3
Simplify the following using index laws, giving your answer as a fully simplified fraction: $\left(\frac{5}{3}\right)^{-2}$(53)−2
Question 4
Simplify the following, giving your answer with positive indices: $\frac{5p^5q^{-4}}{40p^5q^6}$5p5q−440p5q6
Question 5
Simplify $\left(\frac{m^7}{m^{-10}}\right)^2\times\left(\frac{m^5}{m^2}\right)^{-3}$(m7m−10)2×(m5m2)−3, giving your answer with positive indices.