Indices

NZ Level 6 (NZC) Level 1 (NCEA)

Evaluate expressions involving powers

Lesson

We're now going to use what we've covered about powers to evaluate some expressions. If you need to, take some time to revise the multiplication law, division law, power of power rule, zero index law, as well as how to deal with negative indeces.

Here are a few things to remember when evaluating powers.

- Since $b^{-m}=\frac{1}{b^m}$
`b`−`m`=1`b``m`, $b^m$`b``m`will always have the same sign as $b^{-m}$`b`−`m`. - If we have a negative base, because of negatives combining in pairs to make positives, even powers will give a positive result and odd powers will give a negative result.
- The only way to possibly get a result of zero from a power is if the base is zero. Powers of positive or negative numbers can never result in zero.
- $\left(-b\right)^m$(−
`b`)`m`does not mean the same thing as $-b^m$−`b``m`. The first is negative $b$`b`to the power of $m$`m`. The second is the negative of whatever $b$`b`to the power of $m$`m`is. - $\left(\frac{a}{b}\right)^{-m}=\left(\frac{b}{a}\right)^m$(
`a``b`)−`m`=(`b``a`)`m`because the reciprocal of $\frac{a}{b}$`a``b` is $\frac{b}{a}$`b``a`. - If you have to find the power of a mixed fraction, such as $\left(2\frac{1}{4}\right)^5$(214)5, convert it to an improper fraction first.
- Remember that when multiplying decimals, simply perform the whole number multiplication as if the decimal place isn't there, then put the decimal point back. The amount of decimal places in the result is the sum of the amount of decimal places in each decimal being multiplied.
- When you see any irrational number such as $\pi$π, $e$
`e`or $\sqrt{2}$√2, use your calculator when evaluating your final answer and round as specified in the question, or leave the answer in exact form if the question says so.

Is $\left(-3\right)^{-3}$(−3)−3 positive, negative or zero?

Because of the negative index law, we know that $\left(-3\right)^{-3}$(−3)−3 will have the same sign as $\left(-3\right)^3$(−3)3, and a negative base raised to an odd power gives a negative result, hence the result will be negative.

We could have figured this out explicitly as follows, but its often much faster to use the previous method of reasoning when answering this type of question.

$\left(-3\right)^{-3}$(−3)−3 | $=$= | $\frac{1}{\left(-3\right)^3}$1(−3)3 |

$=$= | $\frac{1}{\left(-3\right)\times\left(-3\right)\times\left(-3\right)}$1(−3)×(−3)×(−3) | |

$=$= | $\frac{1}{-27}$1−27 | |

$=$= | $-\frac{1}{27}$−127 |

is $-3^{-3}$−3−3 positive, negative or zero?

$-3^{-3}$−3−3 means the negative of whatever $3^{-3}$3−3 is, so we immediately know that its sign will be negative.

We can evaluate $-3^{-3}$−3−3 as follows.

$-3^{-3}$−3−3 | $=$= | $-\frac{1}{3^3}$−133 |

$=$= | $-\frac{1}{3\times3\times3}$−13×3×3 | |

$=$= | $-\frac{1}{27}$−127 |

Evaluate $\left(6\frac{2}{3}\right)^2$(623)2.

To evaluate $\left(6\frac{2}{3}\right)^2$(623)2 we will first need to convert $6\frac{2}{3}$623 to an improper fraction.

$6$6 is equal to $\frac{18}{3}$183 so $6\frac{2}{3}=\frac{18}{3}+\frac{2}{3}$623=183+23 which equals $\frac{20}{3}$203.

Hence, we can evaluate $\left(6\frac{2}{3}\right)^2$(623)2 as follows.

$\left(6\frac{2}{3}\right)^2$(623)2 | $=$= | $\left(\frac{20}{3}\right)^2$(203)2 |

$=$= | $\frac{20^2}{3^2}$20232 | |

$=$= | $\frac{20\times20}{3\times3}$20×203×3 | |

$=$= | $\frac{400}{9}$4009 |

Evaluate $\left(0.012\right)^2$(0.012)2.

$\left(0.012\right)^2=0.012\times0.012$(0.012)2=0.012×0.012 and we can momentarily ignore the decimal point and perform the whole number multiplication $12\times12$12×12 which equals $144$144.

In $0.012$0.012 there is $3$3 decimal places, so adding $3$3 twice tells us that our result should have $6$6 decimal places, so our answer is $0.000144$0.000144.

Is $\left(\frac{1}{4}\right)^{-2}$(14)−2 positive, negative or $0$0?

$0$0

Apositive

Bnegative

C$0$0

Apositive

Bnegative

C

Find the value of $e^{-2.478}$`e`−2.478 to four decimal places.

Generalise the properties of operations with rational numbers, including the properties of exponents

Apply algebraic procedures in solving problems