1

a

$3$ keer

b

$5$ miljoen

2

a

$2^{6} = 64$

b

Uitgaande van een generatie van 30 jaar, zijn dat (ongeveer) 67 generaties.
Dus $2^{67} \approx 1,5 \cdot 10^{20}$ voorouders

c

Je telt heel veel voorouders dubbel.

3

a

Zie tabel.

b

c

$B (t) = 2^{t}$

d

(ongeveer) $5,7$ , zie figuur.

e

1 milligram = 1000 microgram, dus $2^{t} > 1000$
$2^{9} = 512$ en $2^{10} = 1024$ , dus na $10$ uur

4

a

$2^{8}$

b

$2^{2}$

c

$2^{10}$

5

a

$2^{2} \times 2^{3} = 2^{5}$

b

$2^{p} \times 2^{q} = 2^{p + q}$

c

$2^{p} : 2^{q} = 2^{p - q}$

d

$2^{p} : 2^{p} = 2^{p - p} = 2^{0}$ en duidelijk geldt ook $2^{p} : 2^{p} = 1$ , dus $2^{0} = 1$ .

e

${(2^{3})}^{4} = 2^{12}$

f

${(2^{p})}^{q} = 2^{p \cdot q}$

6

$2^{7} \cdot 2^{5} = 2^{12}$	$2^{7} : 2^{5} = 2^{2}$	${(2^{7})}^{5} = 2^{35}$
$2 \cdot 2^{7} = 2^{8}$	$2^{7} : 2 = 2^{6}$	$1 = 2^{0}$

7

a

$10^{4}$

b

$3^{7} \cdot 2^{7} = 6^{7}$

c

$a^{p} \cdot b^{p} = {(a \cdot b)}^{p}$

d

$\frac{6^{4}}{3^{4}} = \frac{6 \cdot 6 \cdot 6 \cdot 6}{3 \cdot 3 \cdot 3 \cdot 3} = \frac{6}{3} \cdot \frac{6}{3} \cdot \frac{6}{3} \cdot \frac{6}{3} = 2 \cdot 2 \cdot 2 \cdot 2 = 2^{4}$

e

$\frac{a^{p}}{b^{p}} = {(\frac{a}{b})}^{p}$

8

$\frac{x^{5} \cdot x^{3}}{x^{2} \cdot x^{4}} = \frac{x^{8}}{x^{6}} = x^{8 - 6} = x^{2}$	$\frac{{(y^{5})}^{3}}{{(y^{2})}^{4}} = \frac{y^{15}}{y^{8}} = y^{15 - 8} = y^{7}$
$\frac{{(a^{2})}^{5} \cdot a^{3}}{a^{13}} = \frac{a^{13}}{a^{13}} = 1$	$\frac{{(p q)}^{5}}{p^{2} \cdot q^{3}} = \frac{p^{5} \cdot q^{5}}{p^{2} \cdot q^{3}} = p^{5 - 2} \cdot q^{5 - 3} = p^{3} \cdot q^{2}$
$\frac{x^{3} \cdot 64 x^{2}}{{(2 x)}^{5}} = \frac{64 x^{5}}{32 x^{5}} = 2$	$\frac{{(3 y^{2})}^{4}}{81 y^{3}} = \frac{81 y^{8}}{81 y^{3}} = y^{5}$
$\frac{{({(p^{2})}^{3})}^{4}}{p^{2} \cdot p^{3} \cdot p^{4}} = \frac{p^{2 \cdot 3 \cdot 4}}{p^{2 + 3 + 4}} = \frac{p^{24}}{p^{9}} = p^{15}$	$\frac{{(a^{2} b)}^{3}}{a^{5} b} = \frac{a^{6} \cdot b^{3}}{a^{5} \cdot b} = a \cdot b^{2}$
$\frac{{(6 a \cdot b^{2})}^{3}}{2 {(2 a)}^{2} b^{5}} = \frac{6^{3} \cdot a^{3} \cdot b^{6}}{2^{3} \cdot a^{2} \cdot b^{5}} = 3^{3} a b$

9

$32 \cdot 2^{k} = 2^{5} \cdot 2^{k} = 2^{k + 5}$	$2 \cdot 2^{k} = 2^{1} \cdot 2^{k} = 2^{k + 1}$
$2^{k} \cdot 2^{k} = 2^{k + k} = 2^{2 k}$	$8^{k} = {(2^{3})}^{k} = 2^{3 k}$
$16^{k} \cdot 32^{k} = {(2^{4})}^{k} \cdot {(2^{5})}^{k} = 2^{4 k + 5 k} = 2^{9 k}$	$2^{k} \cdot 4^{m} = 2^{k} \cdot {(2^{2})}^{m} = 2^{k + 2 m}$
$\frac{32}{2^{k}} = \frac{2^{5}}{2^{k}} = 2^{5 - k}$	$\frac{2^{k}}{2} = \frac{2^{k}}{2^{1}} = 2^{k - 1}$
$\frac{2^{k + 1}}{2^{k - 1}} = 2^{k + 1 - (k - 1)} = 2^{2}$	$\frac{8^{k}}{8} = \frac{{(2^{3})}^{k}}{2^{3}} = \frac{2^{3 k}}{2^{3}} = 2^{3 k - 3}$
$\frac{32^{k}}{16^{k}} = \frac{2^{5 k}}{2^{4 k}} = 2^{5 k - 4 k} = 2^{k}$	$\frac{4^{m}}{2^{k}} = \frac{{(2^{2})}^{m}}{2^{k}} = \frac{2^{2 m}}{2^{k}} = 2^{2 m - k}$

10

a

onjuist	juist	onjuist
onjuist	juist	juist

b

$3^{n} \cdot 9^{n} = 27^{n}$ of ${(3 \cdot 9)}^{n} = 27^{n}$
$4^{n} : 8^{n} = {(\frac{1}{2})}^{n}$
$2^{n} \times 2^{n} = 2^{2 n}$

11

a

$1,5 \times 1,5 = 2,25 \neq 2$

b

$1,4 \times 1,4 = 1,96 \neq 2$

c

$1,414$

d

Per kwartier: $g^{4} = 2$ $\to$ $g = \sqrt[4]{2} \approx 1,189$
Per 20 minuten: $g^{3} = 2$ $\to$ $g = \sqrt[3]{2} \approx 1,260$

12

a

Als een kolonie in een uur 6 keer zo groot wordt, dan wordt de kolonie in 45 minuten $6^{\frac{3}{4}}$ keer zo groot.

b

$6^{\frac{3}{4}} \approx 3,834$
${3,834}^{4} \approx 216,077$
${(6^{\frac{3}{4}})}^{4} = 6^{\frac{3}{4} \times 4} = 6^{3} = 216$

c

Als een kolonie in een uur 4 keer zo groot wordt, dan wordt de kolonie in 1 uur en 20 minuten $4^{1 \frac{1}{3}}$ keer zo groot.
${(4^{1 \frac{1}{3}})}^{3} = 4^{1 \frac{1}{3} \times 3} = 4^{4} = 256$

13

$1000^{\frac{1}{3}} = {(10^{3})}^{\frac{1}{3}} = 10^{1} = 10$	$1000^{\frac{2}{3}} = {(10^{3})}^{\frac{2}{3}} = 10^{2} = 100$	$16^{\frac{1}{4}} = {(2^{4})}^{\frac{1}{4}} = 2^{1} = 2$
$16^{\frac{3}{4}} = {(2^{4})}^{\frac{3}{4}} = 2^{3} = 8$	$49^{\frac{1}{2}} = {(7^{2})}^{\frac{1}{2}} = 7^{1} = 7$	$49^{1 \frac{1}{2}} = {(7^{2})}^{1 \frac{1}{2}} = 7^{3} = 343$

14

a

$T = 0,2 \cdot {(149,5)}^{1 \frac{1}{2}} \approx 365,5978759$ ;
Dit klopt aardig met de werkelijke omlooptijd van $365,25$ dagen

b

$T = 0,2 \cdot {(1427)}^{1 \frac{1}{2}} \approx 10781,2$ dagen $\approx 29,52$ jaar

15

a

$64^{\frac{1}{2}} = 8$ , want $8^{2} = 64$ ,
$64^{\frac{1}{3}} = 4$ , want $4^{3} = 64$ ,
$64^{\frac{1}{6}} = 2$ , want $2^{6} = 64$ ,
$64^{\frac{5}{6}} = {(64^{\frac{1}{6}})}^{5} = 2^{5} = 32$ .

b

Ze kloppen allemaal!

16

$\frac{{(a^{4})}^{\frac{1}{2}}}{{(a^{\frac{1}{4}})}^{5}} = \frac{a^{2}}{a^{\frac{5}{4}}} = a^{2 - \frac{5}{4}} = a^{\frac{3}{4}}$	$\frac{b^{3} \cdot b^{\frac{1}{3}}}{{(b^{\frac{1}{2}})}^{6}} = \frac{b^{3 \frac{1}{3}}}{b^{3}} = b^{3 \frac{1}{3} - 3} = b^{\frac{1}{3}}$
$\frac{{(c^{12})}^{\frac{2}{3}}}{c^{\frac{1}{2}} \cdot c^{\frac{1}{3}} \cdot c^{\frac{1}{6}}} = \frac{c^{8}}{c^{1}} = c^{7}$	$\frac{{(d^{\frac{4}{3}})}^{\frac{3}{2}}}{{(d^{\frac{2}{5}})}^{\frac{5}{2}}} = \frac{d^{2}}{d^{1}} = d$

17

$\sqrt{a^{3}} = {(a^{3})}^{\frac{1}{2}} = a^{\frac{3}{2}}$	$\sqrt[3]{b^{2}} = {(b^{2})}^{\frac{1}{3}} = b^{\frac{2}{3}}$
$\sqrt{\sqrt{c}} = {(c^{\frac{1}{2}})}^{\frac{1}{2}} = c^{\frac{1}{4}}$	$\sqrt[3]{\sqrt{d^{5}}} = {({(d^{5})}^{\frac{1}{2}})}^{\frac{1}{3}} = d^{\frac{5}{6}}$

18

$x^{2} \sqrt{x} = x^{2} \cdot x^{\frac{1}{2}} = x^{2 \frac{1}{2}}$	$x \sqrt[3]{x} = x^{1} \cdot x^{\frac{1}{3}} = x^{\frac{4}{3}}$
$\sqrt{x \sqrt{x}} = {(x^{1} \cdot x^{\frac{1}{2}})}^{\frac{1}{2}} = {(x^{1 \frac{1}{2}})}^{\frac{1}{2}} = x^{\frac{3}{4}}$	$\frac{x}{\sqrt[3]{x}} = \frac{x^{1}}{x^{\frac{1}{3}}} = x^{1 - \frac{1}{3}} = x^{\frac{2}{3}}$