NZ Level 6 (NZC) Level 1 (NCEA)
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Further expressions involving multiplication law with variable bases
Lesson

In Multiplying algebraic terms with powers, we extended the application of the multiplication law from expressions with numeric bases, like $3^4\times3^6=3^{10}$34×36=310, to expressions with algebraic bases, like $p^8\times p^5=p^{13}$p8×p5=p13. In this lesson we will see how we can apply this idea to larger products, to powers of powers, and to expressions with more complicated algebraic bases.

Exploration

When we come across problems that at first appear quite complicated, we can often make progress toward a solution by breaking up the larger, more difficult problem into a combination of smaller, easier problems that we are more familiar with. Through this process we can find new rules that will help us with similar problems in the future. Soon enough the tricky problems become just as easy as the easy ones!

If we wanted to simplify the expression $b^3\times b^5$b3×b5, we could quickly see that adding the powers gives us the equivalent expression $b^{3+5}=b^8$b3+5=b8. But what if we wanted to simplify a larger product with more factors, like $b^3\times b^5\times b^9$b3×b5×b9? Well we already know that $b^3\times b^5=b^8$b3×b5=b8, so we can first rewrite the expression as

$b^3\times b^5\times b^9=b^8\times b^9$b3×b5×b9=b8×b9.

And now we have a form that is much more familiar, we can use the multiplication law once more to arrive at $b^8\times b^9=b^{17}$b8×b9=b17. Notice that the final power is the sum of all three of the original powers: $3+5+9=17$3+5+9=17. Is there a general pattern we can apply to future problems like this? Yes!

The extended multiplication law

For any base $a$a, and any series of powers $m_1$m1, $m_2$m2, $m_3$m3, $\dots$, up to $m_n$mn,

$a^{m_1}\times a^{m_2}\times a^{m_3}\times\dots\times a^{m_n}=a^{m_1+m_2+m_3+\dots+m_n}$am1×am2×am3××amn=am1+m2+m3++mn

When multiplying any number of terms with the same base, we can keep the base the same and sum all the powers.

In the examples that follow, think about how we can use a combination of simple rules, like the multiplication law together with the power of a power law, to solve challenging problems.

 

Worked example

Write the expression $4y^4\times6y^8\times3y^5$4y4×6y8×3y5 in the simplest index form.

Think: This expression contains numbers and algebraic terms with powers, all multiplied together. We can evaluate the numeric product separately to simplifying the algebraic terms.

Do:

$4y^4\times6y^8\times3y^5$4y4×6y8×3y5 $=$= $4\times6\times8\times y^4\times y^8\times y^5$4×6×8×y4×y8×y5  
  $=$= $192\times y^4\times y^8\times y^5$192×y4×y8×y5 (Evaluate the numeric product first)
  $=$= $192\times y^{4+8+5}$192×y4+8+5 (Using the extended multiplication law)
  $=$= $192y^{17}$192y17  

Reflect: In the first line we were able to rearrange the expression because it doesn't matter what order we multiply terms in a product, the result will stay the same. It can be helpful to rearrange expressions to make it clear what the next step in the simplification will be.

 

Worked example

Simplify the expression $\left(x-6\right)^8\left(x-6\right)^{10}$(x6)8(x6)10.

Think: Both terms in the product, $\left(x-6\right)^8$(x6)8 and $\left(x-6\right)^{10}$(x6)10, have a common base of $\left(x-6\right)$(x6). This means we can use the multiplication law to get a simplified expression in the form $\left(x-6\right)^{\editable{}}$(x6).

Do:

$\left(x-6\right)^8\left(x-6\right)^{10}$(x6)8(x6)10 $=$= $\left(x-6\right)^{8+10}$(x6)8+10
  $=$= $\left(x-6\right)^{18}$(x6)18

Reflect: The use of brackets to specify the base expression is important. This is what lets us tell the difference between something like $\left(x-6\right)^8$(x6)8, where the base is $\left(x-6\right)$(x6) and the power is $8$8, and $x-6^8$x68, where the number $6^8$68 is being subtracted from the variable $x$x.

 

Combining laws

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.

 

Worked example

Simplify the expression $\left(a^2b\right)^3\times\left(ab^2\right)^4$(a2b)3×(ab2)4.

Think: Look at each product separately; we can rewrite $\left(a^2b\right)^3$(a2b)3 using the power of a power law, and similarly for $\left(ab^2\right)^4$(ab2)4. Then we can complete the simplification using the multiplication law.

Considering the first product, let's look at how it is written in expanded form:

$\left(a^2b\right)^3$(a2b)3 $=$= $a^2b\times a^2b\times a^2b$a2b×a2b×a2b
  $=$= $a^2\times a^2\times a^2\times b\times b\times b$a2×a2×a2×b×b×b
  $=$= $a^{2+2+2}\times b^{1+1+1}$a2+2+2×b1+1+1
  $=$= $a^6b^3$a6b3

So in this case we can see that $\left(a^2b\right)^3=a^6b^3$(a2b)3=a6b3. Can you see a pattern here?

Do: Using the idea above,

$\left(a^2b\right)^3\times\left(ab^2\right)^4$(a2b)3×(ab2)4 $=$= $a^{2\times3}b^{1\times3}\times a^{1\times4}b^{2\times4}$a2×3b1×3×a1×4b2×4
  $=$= $a^6b^3\times a^4b^8$a6b3×a4b8
  $=$= $a^{6+4}\times b^{3+8}$a6+4×b3+8
  $=$= $a^{10}b^{11}$a10b11

Reflect: This example featured products of algebraic terms with powers, that themselves were also raised to a power. By writing out the expression in expanded form we have seen how we can use the pattern to generalise the result. That is, when we simplified $\left(a^2b\right)^3=a^6b^3$(a2b)3=a6b3, we multiplied the power of each term within the brackets by the power outside the brackets.

The power of multiple powers

For any base $a$a and $b$b, and for any powers $m$m, $n$n and $p$p,

$\left(a^mb^n\right)^p=a^{m\times p}\times b^{n\times p}$(ambn)p=am×p×bn×p

When applying the power of a power rule to a base with more than one variable, we can multiply each of the original powers by the outside power.

 

Practice questions

Question 1

Simplify the following, giving your answer in index form: $5zxy\times4xzy\times3xyz\times5xy\times4yx$5zxy×4xzy×3xyz×5xy×4yx.

Question 2

Simplify the following, giving your answer in index form: $\left(-5n^3\right)\times m^4\times\left(-5n^3\right)\times m^5$(5n3)×m4×(5n3)×m5.

Question 3

Simplify the following, giving your answer in index form: $10\left(r^7\right)^5\times\left(2rs\right)^4$10(r7)5×(2rs)4.

Outcomes

NA6-2

Extend powers to include integers and fractions.

91026

Apply numeric reasoning in solving problems

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