Chapter 1 Elementary Arithmetic

Section 1.4 Powers and Roots

1.4.2 Calculating using Powers

The following calculation rules allow one to transform and simplify expressions containing powers or roots:

Info 1.4.13

For

a, b \in ℝ, a, b > 0, p, q \in ℚ

, the following exponent rules hold:

a^{p} \cdot b^{p} = (a \cdot b)^{p}, \frac{a^{p}}{b^{p}} = {(\frac{a}{b})}^{p}, a^{p} \cdot a^{q} = a^{p + q}, \frac{a^{p}}{a^{q}} = a^{p - q}, (a^{p})^{q} = a^{p \cdot q} .

Note that, generally,

(a^{p})^{q} \neq a^{p^{q}}

, i.e. multiple powers should be bracketed. For example,

{(2^{3})}^{2} = 8^{2} = 64

, but

2^{(3^{2})} = 2^{9} = 512

.
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Example 1.4.14

Without brackets one reads

a^{p^{q}}

a^{(p^{q})}

, i.e. for example,

\begin{matrix} 2^{3^{4}} & = & 2^{3 \cdot 3 \cdot 3 \cdot 3} = 2^{81} = 2417851639229258349412352 (exponent evaluated first) \\ (2^{3})^{4} & = & 8^{4} = 4096 (bracket evaluated first) . \end{matrix}

Alternatively, we could have used the exponent rules to calculate

(2^{3})^{4} = 2^{(3 \cdot 4)} = 2^{12} = 4096

Exercise 1.4.15

The following expressions can be simplified using the exponent rules:

$3^{3} \cdot 3^{5} \cdot 3^{- 1}$ $=$ .

$3^{3} \cdot 3^{5} \cdot 3^{- 1} = 3^{3 + 5 - 1} = 3^{7} .$
$4^{2} \cdot 3^{2}$ $=$ .

$4^{2} \cdot 3^{2} = (4 \cdot 3)^{2} = 12^{2} .$

However, when comparing powers and roots, one should take care: not only the values, but also the signs of exponent and base control whether the value of the power is large or small:

Example 1.4.16

For a positive base and a negative exponent, the value of the power decreases when increasing the base:

\begin{matrix} 2^{- 1} & = & \frac{1}{2} = 0.5 \\ 3^{- 1} & = & \frac{1}{3} = 0. \overline{3} \\ 4^{- 1} & = & \frac{1}{4} = 0.25 etc. \end{matrix}

For a negative base the sign of the power alternates when increasing the exponent:

\begin{matrix} (- 2)^{1} & = & - 2 \\ (- 2)^{2} & = & 4 \\ (- 2)^{3} & = & - 8 \\ (- 2)^{4} & = & 16 etc. \end{matrix}

Extracting the root (or exponentiation with a positive number smaller than one) decreases a base

> 1

, but increases a base

< 1

\begin{matrix} \sqrt{2} & = & 1.414 \dots < 2 \\ \sqrt{3} & = & 1.732 \dots < 3 \\ \sqrt{0.5} & = & 0.707 \dots > 0.5 \\ \sqrt{0. \overline{3}} & = & 0.577 \dots > 0. \overline{3} etc. \end{matrix}

Exercise 1.4.17

Arrange the following powers in order of size considering the signs of bases and exponents:

2^{3}

2^{- 3}

3^{2}

(- 3)^{2}

(- 3)^{- 2}

3^{\frac{1}{2}}

2^{\frac{1}{3}}

<

<

<

<

<

=

Here, several ways to the correct solution exist. For example, we can raise all terms to the power of

3

(analogous to the calculation in Example 1.2.616). Exponentiation to the power of

3

is allowed since it is an odd number that does not cancel the sign of the exponent. In contrast, exponentiation to the power of

2

would cancel all signs such that the resulting arrangement would be invalid. We have

\begin{matrix} (2^{3})^{3} & = & 2^{3 \cdot 3} = 2^{9} = 512 \\ (2^{- 3})^{3} & = & 2^{(- 3) \cdot 3} = 2^{- 9} = \frac{1}{512} since the exponent is negative \\ (3^{2})^{3} & = & 9^{3} = 81 \cdot 9 = 729 \\ ((- 3)^{2})^{3} & = & 9^{3} = 729 since the exponent is even \\ ((- 3)^{- 2})^{3} & = & \frac{1}{((- 3)^{2})^{3}} = \frac{1}{729} \\ (3^{\frac{1}{2}})^{3} & = & 3^{\frac{3}{2}} > 3 (since the exponent is larger than one) \\ (2^{\frac{1}{3}})^{3} & = & 2 . \end{matrix}

Comparing these values leads to the following arrangement:

(- 3)^{- 2} < 2^{- 3} < 2^{\frac{1}{3}} < 3^{\frac{1}{2}} < 2^{3} < 3^{2} = (- 3)^{2} .

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics