When multiplying a number by itself repeatedly, we are able to use exponent notation to write the expression more simply. Here we are going to look at a rule that allows us simplify products that involve the product of powers.
Consider the expression $a^5\times a^3$a5×a3. Notice that the terms share like bases.
Let's think about what this would look like if we distributed the expression:
We can see that there are eight $a$as being multiplied together, and notice that $8$8 is the sum of the powers in the original expression.
So, in our example above,
$a^5\times a^3$a5×a3 | $=$= | $a^{5+3}$a5+3 |
$=$= | $a^8$a8 |
We can avoid having to write each expression in expanded form by using the product of powers property.
For any base number $a$a, and any numbers $m$m and $n$n as powers,
$a^m\times a^n=a^{m+n}$am×an=am+n
That is, when multiplying terms with a common base:
In other words, when multiplying terms with like bases, we add the powers.
Simplify the following, giving your answer in exponential form: $2^2\cdot2^3$22·23.
Simplify the expression $8y^9\cdot5y^7$8y9·5y7.
$\left(a^2\right)^3$(a2)3 | $=$= | $\left(a^2\right)\times\left(a^2\right)\times\left(a^2\right)$(a2)×(a2)×(a2) |
$=$= | $\left(a\times a\right)\times\left(a\times a\right)\times\left(a\times a\right)$(a×a)×(a×a)×(a×a) | |
$=$= | $a\times a\times a\times a\times a\times a$a×a×a×a×a×a | |
$=$= | $a^6$a6 |
In the expanded form, we can see that we are multiplying six groups of $a$a together. That is, $\left(a^2\right)^3=a^6$(a2)3=a6.
We can confirm this result using the product of powers rule:
We know $\left(a^2\right)\times\left(a^2\right)\times\left(a^2\right)=a^{2+2+2}$(a2)×(a2)×(a2)=a2+2+2 which is equal to $a^6$a6.
We can avoid having to write each expression in expanded form by using the power of a power law.
For any base number $a$a, and any numbers $m$m and $n$n as powers,
$\left(a^m\right)^n=a^{m\times n}$(am)n=am×n
That is, when simplifying a term with a power that itself has a power:
We want to simplify:
$\left(r^2\right)^4$(r2)4
Select the three expressions which are equivalent to $\left(r^2\right)^4$(r2)4:
$r^2\cdot r^4$r2·r4
$\left(r\cdot r\right)\cdot\left(r\cdot r\cdot r\cdot r\right)$(r·r)·(r·r·r·r)
$\left(r\cdot r\right)^4$(r·r)4
$\left(r\cdot r\right)\cdot\left(r\cdot r\right)\cdot\left(r\cdot r\right)\cdot\left(r\cdot r\right)$(r·r)·(r·r)·(r·r)·(r·r)
$r^2\cdot r^2\cdot r^2\cdot r^2$r2·r2·r2·r2
Choose the correct statement:
$\left(r^2\right)^4=r^{2+4}$(r2)4=r2+4
$\left(r^2\right)^4=r^{2\cdot4}$(r2)4=r2·4
Fill in the box to complete the rule: $\left(r^2\right)^4=r^{\editable{}}$(r2)4=r
Express the following in simplified exponential form:
$\left(f^8\right)^6$(f8)6
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 parentheses 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
In other words, a product raised to a power is equivalent to the product of the two factors each raised to the same power.
$\left(-2x^2\right)^3=-8x^6$(−2x2)3=−8x6
Simplify $\left(w^9v^6\right)^4$(w9v6)4.
Simplify, and evaluate where possible, the following expression:
$\left(-3p^2\right)^5$(−3p2)5
In what situations could we use the multiplication law and the power of a power law together? Think about an expression like $\left(u^3\right)^5\times u^4$(u3)5×u4. The first term in the product, $\left(u^3\right)^5$(u3)5, has a base of $\left(u^3\right)$(u3) and a power of $5$5, and the second term, $u^4$u4, has a base of $u$u and a power of $4$4. At this point the two bases are different, so we cannot apply the multiplication law just yet.
One first step could be to rewrite the first term using the power of a power rule:
$\left(u^3\right)^5\times u^4$(u3)5×u4 | $=$= | $u^{3\times5}\times u^4$u3×5×u4 |
$=$= | $u^{15}\times u^4$u15×u4 |
Now both terms have a base of $u$u, so we can combine them by adding the powers: $u^{15+4}=u^{19}$u15+4=u19.
Simplify the expression $8y^3\cdot9y^4$8y3·9y4.
Simplify the following, giving your answer in exponential form: $5u^4v^2\cdot8u^3v^4$5u4v2·8u3v4.
Simplify the following, giving your answer in exponential form: $\left(-8q^4\right)\cdot p^2\cdot\left(-7q^4\right)\cdot p^3$(−8q4)·p2·(−7q4)·p3.