#### 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:

${a}^{p}·{b}^{p}=\left(a·b{\right)}^{p} , \frac{{a}^{p}}{{b}^{p}}={\left(\frac{a}{b}\right)}^{p} , {a}^{p}·{a}^{q}={a}^{p+q} , \frac{{a}^{p}}{{a}^{q}}={a}^{p-q} , \left({a}^{p}{\right)}^{q}={a}^{p·q} .$

Note that, generally, $\left({a}^{p}{\right)}^{q}\ne {a}^{{p}^{q}}$, i.e. multiple powers should be bracketed. For example, ${\left({2}^{3}\right)}^{2}={8}^{2}=64$, but ${2}^{\left({3}^{2}\right)}={2}^{9}=512$.
Formula too small? Double click to enlarge.
##### Example 1.4.14
Without brackets one reads ${a}^{{p}^{q}}$ as ${a}^{\left({p}^{q}\right)}$, i.e. for example,

$\begin{array}{ccc}\multicolumn{1}{c}{{2}^{{3}^{4}}}& =\hfill & {2}^{3·3·3·3}\mathrm{ }=\mathrm{ }{2}^{81}\mathrm{ }=\mathrm{ }2417851639229258349412352\mathrm{ }\mathrm{ }\text{(exponent evaluated first)}\hfill \\ \multicolumn{1}{c}{\left({2}^{3}{\right)}^{4}}& =\hfill & {8}^{4}\mathrm{ }=\mathrm{ }4096\mathrm{ }\mathrm{ }\text{(bracket evaluated first)} .\hfill \end{array}$

Alternatively, we could have used the exponent rules to calculate $\left({2}^{3}{\right)}^{4}={2}^{\left(3·4\right)}={2}^{12}=4096$.

##### Exercise 1.4.15
The following expressions can be simplified using the exponent rules:
1. ${3}^{3}·{3}^{5}·{3}^{-1}$$=$ .

2. ${4}^{2}·{3}^{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:

$\begin{array}{ccc}\multicolumn{1}{c}{{2}^{-1}}& =\hfill & \frac{1}{2}\mathrm{ }=\mathrm{ }0.5 \hfill \\ \multicolumn{1}{c}{{3}^{-1}}& =\hfill & \frac{1}{3}\mathrm{ }=\mathrm{ }0.\stackrel{‾}{3} \hfill \\ \multicolumn{1}{c}{{4}^{-1}}& =\hfill & \frac{1}{4}\mathrm{ }=\mathrm{ }0.25\mathrm{ }\text{etc.}\hfill \end{array}$

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

$\begin{array}{ccc}\multicolumn{1}{c}{\left(-2{\right)}^{1}}& =\hfill & -2\hfill \\ \multicolumn{1}{c}{\left(-2{\right)}^{2}}& =\hfill & 4\hfill \\ \multicolumn{1}{c}{\left(-2{\right)}^{3}}& =\hfill & -8\hfill \\ \multicolumn{1}{c}{\left(-2{\right)}^{4}}& =\hfill & 16\mathrm{ }\text{etc.}\hfill \end{array}$

Extracting the root (or exponentiation with a positive number smaller than one) decreases a base $>1$, but increases a base $<1$:

$\begin{array}{ccc}\multicolumn{1}{c}{\sqrt{2}}& =\hfill & 1.414\dots \mathrm{ }\mathrm{ }<\mathrm{ }\mathrm{ }2\hfill \\ \multicolumn{1}{c}{\sqrt{3}}& =\hfill & 1.732\dots \mathrm{ }\mathrm{ }<\mathrm{ }\mathrm{ }3\hfill \\ \multicolumn{1}{c}{\sqrt{0.5}}& =\hfill & 0.707\dots \mathrm{ }\mathrm{ }>\mathrm{ }\mathrm{ }0.5\hfill \\ \multicolumn{1}{c}{\sqrt{0.\stackrel{‾}{3}}}& =\hfill & 0.577\dots \mathrm{ }\mathrm{ }>\mathrm{ }\mathrm{ }0.\stackrel{‾}{3}\mathrm{ }\text{etc.}\hfill \end{array}$

##### Exercise 1.4.17
Arrange the following powers in order of size considering the signs of bases and exponents: ${2}^{3}$, ${2}^{-3}$, ${3}^{2}$, $\left(-3{\right)}^{2}$, $\left(-3{\right)}^{-2}$, ${3}^{\frac{1}{2}}$, ${2}^{\frac{1}{3}}$:

$<$
$<$
$<$
$<$
$<$
$=$
.