1

a

Zie hieronder.

$2^{{}^{2} \log (3) + {}^{2} \log (7)}$	$=$		Regel 1 toepassen
$2^{{}^{2} \log (3)} \cdot 2^{{}^{2} \log (7)}$	$=$		Regel 5 toepassen
$3 \cdot 7$	$=$	$21$

b

Als $2^{t} = 21$ , dan $t = {}^{2} \log (21)$ .

c

$24$ .

d

$7^{{}^{7} \log (3) + {}^{7} \log (8)}$	$=$		Regel 1 toepassen
$7^{{}^{7} \log (3)} \cdot 7^{{}^{7} \log (8)}$	$=$		Regel 5 toepassen
$3 \cdot 8$	$=$	$24$

2

Zie hieronder.

$2^{{}^{2} \log (9) - {}^{2} \log (3)}$	$=$		Regel 2 toepassen
$2^{{}^{2} \log (9)} : 2^{{}^{2} \log (3)}$	$=$		Regel 5 toepassen
$9 : 3$	$=$	$3$

3

De eerste gelijkheid volgt uit regel 3 en de tweede uit regel 5.

4

a

$3$ , $2$ en $4$ .

b

De eerste gelijkheid volgt uit regel 3.
Verder geldt (regel 5): $10^{\log (2)} = 2$ , en (weer regel 5): $2^{{}^{2} \log (8)} = 8$ .

5

$1,161$	$2,322$
$‐ 1,161$	$2,161$
$‐ 1,161$	$‐ 0,161$

6

$x = \frac{\log (11)}{\log (7)} \approx$ $1,232$	$x = \frac{\log (7)}{\log (11)} \approx$ $0,812$
$x = 1 + \frac{\log (11)}{\log (2)} \approx$ $4,459$	$x = \sqrt{\frac{\log (11)}{\log (2)}} \approx 1,860$ of $x = ‐ 1,860$
$x = {(\frac{\log (111)}{\log (2)})}^{2} \approx$ $46,164$	$x = 2 \cdot \frac{\log (111)}{\log (2)} \approx$ $13,589$
$x =$ $0,000$	$x = \frac{‐ \log (11)}{\log (1,1)} \approx$ $‐ 25,159$

7

a

$K + 0,1 K = 1,1 K$ , dus de groeifactor per jaar is $1,1$ .

b

$K (t) = 5432 \cdot {1,1}^{t}$

c

${1,1}^{t} = \frac{10.000}{5432}$ , dus $t = {}^{1,1} \log (\frac{10.000}{5432}) = 6,403069...$ , dus na $77$ maanden.

8

a

$6 \cdot {1,019}^{12} = 7,52$ miljard

b

${1,019}^{x} = 2 \Leftrightarrow x = \frac{\log (2)}{\log (1,019)} \approx 37$ jaar

9

a

$20$ % minder licht per cm, dus $80$ % blijft over.
De vermenigvuldigingsfactor per cm is $0,8$ en $0,8 \cdot 0,8 = 0,64$ .

b

$100 \cdot {0,8}^{x}$ %

c

${0,8}^{x} = 0,4$ , dus $x = \frac{\log (0,4)}{\log (0,8)} = 4,1$ cm, dus $41$ mm.

10

a

$0,712$	$‐ 0,712$
$‐ 1,222$	$1,222$

b

$6,658$	$2,219$
$‐ 0,358$	$‐ 0,119$

11

a

${}^{5} \log (625) + {}^{5} \log (\frac{1}{5})$ $=$ $4 + ‐ 1 = 3$ .
${}^{5} \log (625) + {}^{5} \log (\frac{1}{5})$ $=$ ${}^{5} \log (125)$ $= 3$ .

b

$\log (20) + \log (5) = 1,3 + 0,7 = 2$
$\log (20) + \log (5) = \log (100) = 2$ en
$\log (5) - \log (\frac{1}{2}) = 0,7 - ‐ 0,3 = 1$
$\log (5) - \log (\frac{1}{2}) = \log (10) = 1$

12

$= {}^{3} \log (9) = 2$	$= {}^{5} \log (\frac{1}{5}) = ‐ 1$
$= {}^{\frac{1}{4}} \log (4) = ‐ 1$	$= {}^{30} \log (30) = 1$
$= {}^{2} \log (\frac{1}{4}) = ‐ 2$	$= {}^{2} \log (1) = 0$

13

$= {}^{4} \log (64) = 3$	$= {}^{5} \log (5) = 1$
$= {}^{0,7} \log (\frac{1}{0,7}) = ‐ 1$	$= {}^{7} \log (\frac{84 x}{12 x}) = {}^{7} \log (7) = 1$
$= {}^{5} \log (\frac{6}{5 \cdot 3 \cdot 2}) = {}^{5} \log (\frac{1}{5}) = ‐ 1$	$= \log (\frac{12 \cdot 3 \cdot 125}{45}) = 2$

14

a

In beide gevallen vind je $1,5$ respectievelijk $3,38039...$ .

b

${}^{4} \log (2^{11}) = 11 \cdot {}^{4} \log (2) = 11 \cdot \frac{1}{2} = 5 \frac{1}{2}$
${}^{\frac{1}{4}} \log (2^{11}) = 11 \cdot {}^{\frac{1}{4}} \log (2) = 11 \cdot ‐ \frac{1}{2} = ‐ 5 \frac{1}{2}$
${}^{3} \log ({(\frac{1}{9})}^{11}) = 11 \cdot {}^{3} \log (\frac{1}{9}) = 11 \cdot ‐ 2 = ‐ 22$
${}^{\frac{1}{3}} \log ({(\frac{1}{9})}^{11}) = 11 \cdot {}^{\frac{1}{3}} \log (\frac{1}{9}) = 11 \cdot 2 = 22$

15

$\log (20) = \log (2) + \log (10) = 1,3$ ; $\log (0,05) = \log (\frac{1}{20}) = ‐ 1,3$ ; $\log (0,2) = \log (2) - \log (10) = ‐ 0,7$
$\log (4) = 2 \log (2) = 0,6$ ; $\log (16) = 4 \log (2) = 1,2$ ; $\log (160) = \log (16) + \log (10) = 2,2$
$\log (25) = \log (100) - 2 \log (2) = 1,4$ ; ${}^{2} \log (10) = \frac{\log (10)}{\log (2)} = 3,33$ ; ${}^{5} \log (10) = \frac{\log (10)}{\log (5)} = 1,43$ , want $\log (5) = \log (10) - \log (2)$ .