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CanadaON
Grade 11

Power of a power with integer and variable bases

Lesson

We introduced the power of a power when we learnt about the zero exponent

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)

In the expanded form, we can see that $5$5 is multiplied by itself $6$6 times. That is $\left(5\times5\right)\times\left(5\times5\right)\times\left(5\times5\right)=5^6$(5×5)×(5×5)×(5×5)=56.

To avoid having to write every power of a power expression in expanded form, we want a shortcut in simplifying $\left(5^2\right)^3$(52)3 to get $5^6$56. The shortcut? We MULTIPLY the powers.

Generalising this, we get:

Power of a Power

$\left(x^a\right)^b=x^{a\times b}$(xa)b=xa×b

 

Examples

Question 1

Express in simplified exponential form: $\left(3^5\right)^3$(35)3

Solution:

$\left(3^5\right)^3$(35)3 $=$= $3^{5\times3}$35×3
  $=$= $3^{15}$315

 

Question 2

Simplify using the exponent laws:

$\left(3^5\right)^3\times\left(3^2\right)^3$(35)3×(32)3

 
Question 3

Simplify and evaluate using the exponent laws: $\left(4^2\right)^3\div\left(4^4\right)^0$(42)3÷​(44)0

ThinkWe will have to use the Power of a Power, Zero Exponent and Division Laws.

Starting with the numerator

$\left(4^2\right)^3$(42)3 $=$= $4^{2\times3}$42×3
  $=$= $4^6$46

Now the denominator

$\left(4^4\right)^0$(44)0 $=$= $4^{4\times0}$44×0
  $=$= $4^0$40
  $=$= $1$1

Combining the two:

$\left(4^2\right)^3\div\left(4^4\right)^0$(42)3÷​(44)0 $=$= $\frac{4^6}{1}$461
  $=$= $4^6$46

 

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