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VCE 11 Methods 2023

7.01 Index laws

Lesson

 

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.

 

The index laws
  • 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=amn

  • 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 a0

  • Raising a non-zero number to a negative power results in the reciprocal of the number raised to a positive power.

$a^{-n}=\frac{1}{a^n}\text{for }a\ne0$an=1anfor a0

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. Think about what this would look like if the expression was expanded:

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

To simplify the expression $a^6\div a^2$a6÷​a2, write it in expanded form as:

Dividing top and bottom by $a$a twice leaves $a^4$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
  $=$= $\frac{a^{5x-1}}{2}$a5x12

 

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, notice that $5$5 is multiplied by itself $6$6 times, which was the $5^2$52 multiplied by itself $3$3 times. So, this gave the total index as $3\times2$3×2.

 

Product to a power

To simplify the expression $\left(ab\right)^4$(ab)4, 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

Once expanded, it is then possible to 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

To simplify the expression $\left(\frac{2}{x}\right)^3$(2x)3, 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

Once expanded, 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

When dividing powers with like bases, subtract the indices. What happens when doing this leaves a power of zero? 

$\frac{4^1}{4^1}$4141 $=$= $4^{1-1}$411
  $=$= $4^0$40

 

Notice that the left-hand side simplifies to $1$1. This will also be the case with $\frac{4^2}{4^2}$4242 or any expression 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 the 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

Remember that when dividing powers with like bases, subtract the indices. That is:

$a^x\div a^y=a^{x-y}$ax÷​ay=axy

What happens when $y$y is bigger than $x$x? For example, simplifying $a^3\div a^5$a3÷​a5 using the division law gives $a^{-2}$a2. So, what does a negative index mean? Expand this example to find out:

Remember that when simplifying fractions, divide out common factors in the numerator and denominator. Remember that any number divided by itself is $1$1.

So, using the second approach,  $a^3\div a^5$a3÷​a5 can be expressed with a positive index as $\frac{1}{a^2}$1a2.

 

Negative powers and fractions

Combine the previous 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}}$3a4b6a7b3c4, 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}}$3a4b6a7b3c4 $=$= $\frac{1a^4b}{2a^7b^{-3}c^{-4}}$1a4b2a7b3c4
  $=$= $\frac{a^{\left(4-7\right)}b^{\left(1+3\right)}}{2c^{-4}}$a(47)b(1+3)2c4
  $=$= $\frac{a^{-3}b^4}{2c^{-4}}$a3b42c4

Lastly to write this without negative fractions, we could write these as a product of the components:

$\frac{a^{-3}b^4}{2c^{-4}}$a3b42c4 $=$= $\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}}$3a4b6a7b3c4 $=$= $\frac{1a^4b^1c^0}{2a^7b^{-3}c^{-4}}$1a4b1c02a7b3c4
  $=$= $\frac{b^{\left(1+3\right)}c^{\left(0+4\right)}}{2a^{\left(7-4\right)}}$b(1+3)c(0+4)2a(74)
  $=$= $\frac{b^4c^4}{2a^3}$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}$5p5q440p5q6

Question 5

Simplify $\left(\frac{m^7}{m^{-10}}\right)^2\times\left(\frac{m^5}{m^2}\right)^{-3}$(m7m10)2×(m5m2)3, giving your answer with positive indices.

Outcomes

U1.AoS2.11

the exponent laws

U1.AoS2.2

substitution into, and manipulation of, these expressions

U1.AoS2.3

recognition of equivalent expressions and simplification of algebraic expressions involving different forms of polynomial and power functions, the use of distributive and exponent laws applied to these functions, and manipulation from one form of expression to an equivalent form

U1.AoS2.18

apply distributive and exponent laws to manipulate and simplify expressions involving polynomial and power function, by hand in simple cases

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