$a^m\times a^n=a^{m+n}$am×an=am+n
which tells us that when multiplying terms with like bases we can equivalently add their exponents (or powers). In that lesson we used the multiplication law to simplify terms with numeric bases, like so: $8^5\times8^9=8^{5+9}=8^{14}$85×89=85+9=814.
Now we are ready to apply the same idea to algebraic terms with powers. These will be terms such as $b^7$b7, $k^2$k2, $p^{19}$p19, and so on, where the base is a variable and the power is a whole number.
Simplify the expression $u^{12}\times u^4$u12×u4.
Think: This expression is made up of two terms, $u^{12}$u12 and $u^4$u4, that are being multiplied together. In each case the base is the variable $u$u, so we can simplify the expression using the multiplication law.
Do:
$u^{12}\times u^4$u12×u4 | $=$= | $u^{12+4}$u12+4 |
$=$= | $u^{16}$u16 |
Write the expression $m^6\times8m^5$m6×8m5 in simplified exponential form.
Think: In this expression we have two algebraic terms, $m^6$m6 and $m^5$m5, and one other numeric factor, $8$8. Notice that the number $8$8 has no powers acting upon it. That is, $8m^5=8\times m\times m\times m\times m\times m$8m5=8×m×m×m×m×m.
Do:
$m^6\times8m^5$m6×8m5 | $=$= | $8\times m^6\times m^5$8×m6×m5 |
$=$= | $8\times m^{6+5}$8×m6+5 | |
$=$= | $8m^{11}$8m11 |
Reflect: It may be helpful to insert the explicit multiplication symbol $\times$× between each factor of a product to make it clear what power is acting upon what base. This was done in the first line of the working above.
In the example above we identified that the term $8m^5$8m5 is made up of the multiplier $8$8 and the multiplier $m^5$m5, which itself consists of a base $m$m and a power of $5$5. But what if we want to convey that $8m$8m is the base, and the whole thing is being raised to some power? We can use brackets to make this more clear, as outlined below.
Let's think about the expression $5\times5\times5\times5\times a\times a\times a\times a$5×5×5×5×a×a×a×a. How can we write this in a more compact form, using our knowledge of exponent laws? Firstly, we can see that there are four lots of the number $5$5 and four lots of the variable $a$a being multiplied together. We can simplify the two parts separately like so
$5\times5\times5\times5\times a\times a\times a\times a=5^4\times a^4$5×5×5×5×a×a×a×a=54×a4.
That is a good start, but there is another way we can approach this. Using the fact that multiplication is commutative (the order that we multiply the numbers doesn't change the result) we can rearrange the product to get $5\times a\times5\times a\times5\times a\times5\times a$5×a×5×a×5×a×5×a. Notice that this expression has the same number of $5$5s and $a$as, just in a different order. Now if we look at groups of $5\times a$5×a, we can treat each one as a separate base and simplify the expression in the following way.
$5\times a\times5\times a\times5\times a\times5\times a$5×a×5×a×5×a×5×a | $=$= | $\left(5\times a\right)\times\left(5\times a\right)\times\left(5\times a\right)\times\left(5\times a\right)$(5×a)×(5×a)×(5×a)×(5×a) |
$=$= | $\left(5\times a\right)^4$(5×a)4 |
Using two different approaches, we have seen that $5\times5\times5\times5\times a\times a\times a\times a$5×5×5×5×a×a×a×a can be written as either $5^4\times a^4$54×a4 or $\left(5\times a\right)^4$(5×a)4, and this must mean that $5^4a^4=\left(5a\right)^4$54a4=(5a)4. In the second case, the base is "$5a$5a" and the power is $4$4.
As you might expect, we can use this specific example to arrive at a general rule about bases that are products of two numbers, or two variables, or a number and a variable.
For the product of any numbers $a$a and $b$b in the base, and for any number $n$n in the power,
$\left(ab\right)^n=a^nb^n$(ab)n=anbn
A product raised to a power is equivalent to the product of the two factors each raised to the same power.
Simplify the expression $k^6\times\left(4k\right)^3$k6×(4k)3.
Think: In one term we have a base of $k$k and a power of $6$6, and in the other we have a base of $4k$4k and a power of $3$3. The bases are currently different, so we cannot simplify by adding powers just yet. We need to rewrite $\left(4k\right)^3$(4k)3 first.
Do:
$k^6\times\left(4k\right)^3$k6×(4k)3 | $=$= | $k^6\times4^3\times k^3$k6×43×k3 | (Using the fact that $\left(ab\right)^n=a^nb^n$(ab)n=anbn) |
$=$= | $4^3\times k^6\times k^3$43×k6×k3 | ||
$=$= | $4^3\times k^{6+3}$43×k6+3 | (Using the fact that $a^m\times a^n=a^{m+n}$am×an=am+n) | |
$=$= | $4^3\times k^9$43×k9 | ||
$=$= | $64k^9$64k9 |
Reflect: We were able to expand the term $\left(4k\right)^3$(4k)3 into two more terms: one with a numeric base and the other with a variable base. That this is possible goes back to the fact that we can rearrange the factors in an expanded product any way that we want.
If we have an expression that consists of more than one base, we need to apply the multiplication law separately to each set of like bases.
In an expression such as $x^2y\times xy^2$x2y×xy2, we would simplify the powers of $x^2\times x$x2×x separately to $y\times y^2$y×y2.
It may be useful to split up and rearrange the multiplication to make it obvious which terms have like bases and can therefore be simplified. For example, we may like to rewrite $6ab\times a^2$6ab×a2 as $6\times b\times a\times a^2$6×b×a×a2 to identify that only the powers of $a\times a^2$a×a2 can be added.
Simplify the expression $8x^3y^2\times4x^5y^7$8x3y2×4x5y7.
Think: We will multiply the coefficients first, then simplify the powers of $x$x separately from the powers of $y$y.
Do:$8x^3y^2\times4x^5y^7$8x3y2×4x5y7 | $=$= | $32x^3y^2\times x^5y^7$32x3y2×x5y7 |
$=$= | $32x^8y^2\times y^7$32x8y2×y7 | |
$=$= | $32x^8y^9$32x8y9 |
Simplify the following, giving your answer in exponential form: $y^2\times y^6$y2×y6.
Simplify the expression $4y^9\times6y^2$4y9×6y2.
Simplify the following, giving your answer in exponential form: $7u^3v^2\times4u^2v^4$7u3v2×4u2v4.
Derive, through the investigation and examination of patterns, the exponent rules for multiplying and dividing monomials, and apply these rules in expressions involving one and two variables with positive exponents