1.6 Eksponensiale

Siyavula textbooks: wiskunde Page 1 / 1

Inleiding

In hierdie hoofstuk sal jy leer van 'n eenvoudiger manier om uitdrukkings soos $2 \times 2 \times 2 \times 2$ te skryf. Dit staan bekend as eksponensiaalnotasie .

Definisie

Eksponensiaalnotasie is 'n kort manier om te skryf dat 'n getal meermale met homself vermenigvuldig word. Byvoorbeeld, eerder as om te skryf $5 \times 5 \times 5$ , gebruik ons $5^{3}$ om aan te dui dat die getal 5 drie maal met homself vermenigvuldig word en 'n mens sê "5 tot die mag 3". Soortgelyk is $5^{2}$ dieselfde as $5 \times 5$ en $3^{5}$ is $3 \times 3 \times 3 \times 3 \times 3$ . Laat ons beter definieer hoe om eksponensiaalnotasie te gebruik.

Eksponensiaalnotasie

Eksponensiaalnotasie verwys na 'n getal wat geskryf word as

a^{n}

waar $n$ 'n heelgetal is en $a$ enige reële getal is. Ons noem $a$ die grondtal en $n$ die eksponent .

$a$ tot die mag $n$ is

a^{n} = a \times a \times \dots \times a (n -keer)

Dit wil sê, $a$ word $n$ keer met homself vermenigvuldig.

Ons kan ook 'n negatiewe eksponent, $- n$ , gebruik. In hierdie geval

a^{- n} = \frac{1}{a \times a \times \dots \times a (n -keer)}

Eksponente

Indien $n$ 'n ewe getal is, sal $a^{n}$ altyd 'n positiewe getal wees vir enige reële getal $a$ , behalwe $0$ . Byvoorbeeld, hoewel $- 2$ negatief is, is beide ${(- 2)}^{2} = - 2 \times - 2 = 4$ en ${(- 2)}^{- 2} = \frac{1}{- 2 \times - 2} = \frac{1}{4}$ positief.

Khan academy video oor eksponente 1 (in engels)

Khan academy video oor eksponente 2 (in engels)

Eksponentwette

Daar is heelwat eksponentwette wat ons kan gebruik om getalle met eksponente te vereenvoudig. Sommige van hierdie wette het ons reeds in vorige grade teëgekom, maar ons sal die volledige lys hier sien en elke wet verduidelik, sodat jy hulle kan verstaan en nie bloot memoriseer nie.

\begin{matrix} a^{0} & = & 1 \\ a^{m} \times a^{n} & = & a^{m + n} \\ a^{- n} & = & \frac{1}{a^{n}} \\ a^{m} \div a^{n} & = & a^{m - n} \\ {(a b)}^{n} & = & a^{n} b^{n} \\ {(a^{m})}^{n} & = & a^{m n} \end{matrix}

Eksponente, wet 1: $a^{0} = 1$

Volgens die definisie van eksponensiaalnotasie is

\begin{matrix} a^{0} & = & 1, (a \neq 0) \end{matrix}

Byvoorbeeld, $x^{0} = 1$ en ${(1 000 000)}^{0} = 1$

Toepassing van wet 1: $a^{0} = 1, (a \neq 0)$

$16^{0}$
$16 a^{0}$
${(16 + a)}^{0}$
${(- 16)}^{0}$
$- 16^{0}$

Eksponente, wet 2: $a^{m} \times a^{n} = a^{m + n}$

Khan academy video oor eksponente 3 (in engels)

Die definisie van eksponensiaalnotasie wys dat

\begin{matrix} a^{m} \times a^{n} & = & 1 \times a \times ... \times a & (m -keer) \\ \times 1 \times a \times ... \times a & (n -keer) \\ = & 1 \times a \times ... \times a & (m + n -keer) \\ = & a^{m + n} \end{matrix}

Byvoorbeeld,

\begin{matrix} 2^{7} \times 2^{3} & = & (2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2) \\ = & 2^{7 + 3} \\ = & 2^{10} \end{matrix}

Interessante feit

Hierdie eenvoudige wet is die rede waarom eksponente oorspronklik geskep is. Voor die dae van rekenaars moes vermenigvuldiging met potlood en papier gedoen word. Dit vat baie lank om vermenigvuldiging te doen, maar dit is vinnig en eenvoudig om getalle bymekaar te tel. Hierdie eksponentwet wys dat dit moontlik is om twee getalle te vermenigvuldig deur hulle eksponente bymekaar te tel (indien hulle dieselfde grondtal het). Hierdie ontdekking het wiskundiges baie tyd gespaar, wat hulle toe kon gebruik om iets meer produktiefs te doen.

Toepassing van wet 2: $a^{m} \times a^{n} = a^{m + n}$

$x^{2} \cdot x^{5}$
$2^{3} . 2^{4}$ [Neem kennis dat die grondtal (2) dieselfde bly.]
$3 \times 3^{2 a} \times 3^{2}$

Eksponente, wet 3: $a^{- n} = \frac{1}{a^{n}}, a \neq 0$

Die definisie van eksponensiaalnotasie vir 'n negatiewe eksponent wys dat

\begin{matrix} a^{- n} & = & 1 \div a \div ... \div a & (n -keer) \\ = & \frac{1}{1 \times a \times \dots \times a} & (n -keer) \\ = & \frac{1}{a^{n}} \end{matrix}

Dit beteken dat 'n minus teken in die eksponent 'n alternatiewe manier is om aan te dui dat die hele eksponensiaal gedeel eerder asvermenigvuldig moet word.

Byvoorbeeld,

\begin{matrix} 2^{- 7} & = & \frac{1}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} \\ = & \frac{1}{2^{7}} \end{matrix}

Toepassing van wet 3: $a^{- n} = \frac{1}{a^{n}}, a \neq 0$

$2^{- 2} = \frac{1}{2^{2}}$
$\frac{2^{- 2}}{3^{2}}$
${(\frac{2}{3})}^{- 3}$
$\frac{m}{n^{- 4}}$
$\frac{a^{- 3} \cdot x^{4}}{a^{5} \cdot x^{- 2}}$

Eksponente, wet 4: $a^{m} \div a^{n} = a^{m - n}$

Met Wet 3 het ons reeds besef dat 'n minusteken 'n manier is om te wys dat die eksponensiaal gedeel eerder as vermenigvuldig moetword. Wet 4 is basies 'n meer algemene manier om dieselfde stelling te maak. Ons verkry hierdie wet deur Wet 3 aan beide kante met $a^{m}$ te vermenigvuldig en dan Wet 2 te gebruik.

\begin{matrix} \frac{a^{m}}{a^{n}} & = & a^{m} a^{- n} \\ = & a^{m - n} \end{matrix}

Byvoorbeeld,

\begin{matrix} 2^{7} \div 2^{3} & = & \frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}{2 \times 2 \times 2} \\ = & 2 \times 2 \times 2 \times 2 \\ = & 2^{4} \\ = & 2^{7 - 3} \end{matrix}

Khan academy video oor eksponente 4 (in engels)

Toepassing van wet 4: $a^{m} \div a^{n} = a^{m - n}$

$\frac{a^{6}}{a^{2}} = a^{6 - 2}$
$\frac{3^{2}}{3^{6}}$
$\frac{32 a^{2}}{4 a^{8}}$
$\frac{a^{3 x}}{a^{4}}$

Eksponente, wet 5: ${(a b)}^{n} = a^{n} b^{n}$

Die volgorde waarin twee getalle vermenigvuldig word, is onbelangrik. Dus,

\begin{matrix} {(a b)}^{n} & = & a \times b \times a \times b \times ... \times a \times b & (n -keer) \\ = & a \times a \times ... \times a & (n -keer) \\ \times b \times b \times ... \times b & (n -keer) \\ = & a^{n} b^{n} \end{matrix}

Byvoorbeeld,

\begin{matrix} {(2 \cdot 3)}^{4} & = & (2 \cdot 3) \times (2 \cdot 3) \times (2 \cdot 3) \times (2 \cdot 3) \\ = & (2 \times 2 \times 2 \times 2) \times (3 \times 3 \times 3 \times 3) \\ = & (2^{4}) \times (3^{4}) \\ = & 2^{4} 3^{4} \end{matrix}

Toepassing van wet 5: ${(a b)}^{n} = a^{n} b^{n}$

${(2 x y)}^{3} = 2^{3} x^{3} y^{3}$
${(\frac{7 a}{b})}^{2}$
${(5 a)}^{3}$

Eksponente, wet 6: ${(a^{m})}^{n} = a^{m n}$

Dit is moontlik om die eksponensiaal van 'n eksponensiaal te bereken. Die eksponensiaal van 'n getal is 'n reële getal. So, selfs al klink die eerste sin ingewikkeld, beteken dit bloot dat 'n mens die eksponensiaal van 'n getal bereken en dan die eksponensiaal van die resultaat bereken.

\begin{matrix} {(a^{m})}^{n} & = & a^{m} \times a^{m} \times ... \times a^{m} & (n -keer) \\ = & a \times a \times ... \times a & (m \times n -keer) \\ = & a^{m n} \end{matrix}

Byvoorbeeld,

\begin{matrix} {(2^{2})}^{3} & = & (2^{2}) \times (2^{2}) \times (2^{2}) \\ = & (2 \times 2) \times (2 \times 2) \times (2 \times 2) \\ = & (2^{6}) \\ = & 2^{(2 \times 3)} \end{matrix}

Toepassing van wet 6: ${(a^{m})}^{n} = a^{m n}$

${(x^{3})}^{4}$
${[{(a^{4})}^{3}]}^{2}$
${(3^{n + 3})}^{2}$

Vereenvoudig: $\frac{5^{2 x - 1} \cdot 9^{x - 2}}{15^{2 x - 3}}$

Faktoriseer al die grondtalle in priemfaktore
$\begin{matrix} = & \frac{5^{2 x - 1} \cdot {(3^{2})}^{x - 2}}{{(5.3)}^{2 x - 3}} \\ = & \frac{5^{2 x - 1} \cdot 3^{2 x - 4}}{5^{2 x - 3} \cdot 3^{2 x - 3}} \end{matrix}$
Tel die eksponente van ooreenstemmende grondtalle bymekaar, volgens Wette 2 en 4
$\begin{matrix} = & 5^{2 x - 1 - 2 x + 3} \cdot 3^{2 x - 4 - 2 x + 3} \\ = & 5^{2} \cdot 3^{- 1} \end{matrix}$
Skryf die vereenvoudigde antwoord met positiewe eksponente
$= \frac{25}{3}$

Ondersoek: eksponensiale

Skryf die korrekte antwoord in the Antwoord kolom. Die beskikbare antwoorde is: $\frac{3}{2}$ , 1, $- 1$ , $- \frac{1}{3}$ , 8. Antwoorde mag herhaal word.

Vraag	Antwoord
$2^{3}$
$7^{3 - 3}$
${(\frac{2}{3})}^{- 1}$
$8^{7 - 6}$
${(- 3)}^{- 1}$
${(- 1)}^{23}$

Die volgende video gee 'n voorbeeld van hoe om sommige van die konsepte wat in hierdie hoofstuk gedek is, te gebruik.

Khan academy video oor eksponente 5 (in engels)

Hoofstukoefeninge

Vereenvoudig so ver as moontlik.
1. $302^{0}$
2. $1^{0}$
3. ${(x y z)}^{0}$
4. ${[{(3 x^{4} y^{7} z^{12})}^{5} {(- 5 x^{9} y^{3} z^{4})}^{2}]}^{0}$
5. ${(2 x)}^{3}$
6. ${(- 2 x)}^{3}$
7. ${(2 x)}^{4}$
8. ${(- 2 x)}^{4}$
Vereenvoudig sonder om 'n sakrekenaar te gebruik. Skryf antwoorde met positiewe eksponente.
1. $\frac{3 x^{- 3}}{{(3 x)}^{2}}$
2. $5 x^{0} + 8^{- 2} - {(\frac{1}{2})}^{- 2} \cdot 1^{x}$
3. $\frac{5^{b - 3}}{5^{b + 1}}$
Vereenvoudig en wys alle stappe.
1. $\frac{2^{a - 2} . 3^{a + 3}}{6^{a}}$
2. $\frac{a^{2 m + n + p}}{a^{m + n + p} \cdot a^{m}}$
3. $\frac{3^{n} \cdot 9^{n - 3}}{27^{n - 1}}$
4. ${(\frac{2 x^{2 a}}{y^{- b}})}^{3}$
5. $\frac{2^{3 x - 1} \cdot 8^{x + 1}}{4^{2 x - 2}}$
6. $\frac{6^{2 x} \cdot 11^{2 x}}{22^{2 x - 1} \cdot 3^{2 x}}$
Vereenvoudig sonder om 'n sakrekenaar te gebruik.
1. $\frac{{(- 3)}^{- 3} \cdot {(- 3)}^{2}}{{(- 3)}^{- 4}}$
2. ${(3^{- 1} + 2^{- 1})}^{- 1}$
3. $\frac{9^{n - 1} \cdot 27^{3 - 2 n}}{81^{2 - n}}$
4. $\frac{2^{3 n + 2} \cdot 8^{n - 3}}{4^{3 n - 2}}$

Questions & Answers

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In economics, a perfect market refers to a theoretical construct where all participants have perfect information, goods are homogenous, there are no barriers to entry or exit, and prices are determined solely by supply and demand. It's an idealized model used for analysis,

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When MP₁ becomes negative, TP start to decline. Extuples Suppose that the short-run production function of certain cut-flower firm is given by: Q=4KL-0.6K2 - 0.112 • Where is quantity of cut flower produced, I is labour input and K is fixed capital input (K-5). Determine the average product of lab

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Answer

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Source: OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4

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1.6 Eksponensiale

Inleiding

Definisie

Khan academy video oor eksponente 1 (in engels)

Khan academy video oor eksponente 2 (in engels)

Eksponentwette

Eksponente, wet 1: a 0 = 1

Toepassing van wet 1: a 0 = 1 , ( a ≠ 0 )

Eksponente, wet 2: a m × a n = a m + n

Khan academy video oor eksponente 3 (in engels)

Interessante feit

Toepassing van wet 2: a m × a n = a m + n

Eksponente, wet 3: a - n = 1 a n , a ≠ 0

Toepassing van wet 3: a - n = 1 a n , a ≠ 0

Eksponente, wet 4: a m ÷ a n = a m - n

Khan academy video oor eksponente 4 (in engels)

Toepassing van wet 4: a m ÷ a n = a m - n

Eksponente, wet 5: ( a b ) n = a n b n

Toepassing van wet 5: ( a b ) n = a n b n

Eksponente, wet 6: ( a m ) n = a m n

Toepassing van wet 6: ( a m ) n = a m n

Ondersoek: eksponensiale

Khan academy video oor eksponente 5 (in engels)

Hoofstukoefeninge

Questions & Answers

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Eksponente, wet 1: $a^{0} = 1$

Toepassing van wet 1: $a^{0} = 1, (a \neq 0)$

Eksponente, wet 2: $a^{m} \times a^{n} = a^{m + n}$

Toepassing van wet 2: $a^{m} \times a^{n} = a^{m + n}$

Eksponente, wet 3: $a^{- n} = \frac{1}{a^{n}}, a \neq 0$

Toepassing van wet 3: $a^{- n} = \frac{1}{a^{n}}, a \neq 0$

Eksponente, wet 4: $a^{m} \div a^{n} = a^{m - n}$

Toepassing van wet 4: $a^{m} \div a^{n} = a^{m - n}$

Eksponente, wet 5: ${(a b)}^{n} = a^{n} b^{n}$

Toepassing van wet 5: ${(a b)}^{n} = a^{n} b^{n}$

Eksponente, wet 6: ${(a^{m})}^{n} = a^{m n}$

Toepassing van wet 6: ${(a^{m})}^{n} = a^{m n}$