1

a

$x = 2^{3} = 8$	$x = 7$
$x = 4$	$x = 3 \frac{1}{2}$
$x = 64$	$x = 3$ of $x = ‐ 3$
$x = \frac{1}{8}$	$x = 4$ of $x = ‐ 2$

b

$log (x) = 3 log (6)$	$3 log (x) = log (6)$
$log (x) = log (6^{3})$	$log (x^{3}) = log (6)$
$x = 216$	$x^{3} = 6$
	$x = \sqrt[3]{6}$

$2 log (\frac{1}{x}) = 3 log (4)$	$log (6) + log (\frac{1}{x}) = log (x)$
$log (\frac{1}{x}) = \frac{3}{2} log (4)$	$log (\frac{6}{x}) = log (x)$
$log (\frac{1}{x}) = log (4^{\frac{3}{2}})$	$\frac{6}{x} = x$
$x = 2^{‐3} = \frac{1}{8}$	$x = \sqrt{6}$ (Niet $x = ‐ \sqrt{6}$ , want $x > 0$ )

c

${}^{2} log (x) + {}^{2} log (8) = {}^{2} log (12)$	${}^{2} log (95) - {}^{2} log (x) = {}^{2} log (5)$
${}^{2} log (8 x) = {}^{2} log (12)$	${}^{2} log (\frac{95}{x}) = {}^{2} log (5)$
$8 x = 12 \Leftrightarrow x = 1 \frac{1}{2}$	$\frac{95}{x} = 5 \Leftrightarrow x = 19$

${}^{5} log (x) - {}^{5} log (2) = {}^{5} log (7)$	$log (x) + log (40) = 4$
${}^{5} log (\frac{x}{2}) = {}^{5} log (7)$	$log (40 x) = 4$
$\frac{x}{2} = 7 \Leftrightarrow x = 14$	$40 x = 10^{4} \Leftrightarrow x = 250$

$log (x) - log (5) = log (70)$	${}^{2} log (x) - {}^{2} log (3) = {}^{2} log (12) - {}^{2} log (x)$
$log (\frac{x}{5}) = log (70)$	${}^{2} log (\frac{x}{3}) = {}^{2} log (\frac{12}{x})$
$x = 350$	$\frac{x}{3} = \frac{12}{x} \Leftrightarrow x^{2} = 36$
	Alleen $x = 6$ voldoet.

2

a

Domein $f$ : alle getallen $x$ met $x \neq 0$
Domein $g$ : de getallen $x$ met $x > 0$
Domein $h$ : de getallen $x$ met $x > ‐ 6$

b

Regel 3, maar die geldt alleen voor positieve $x$ , dus het deel rechts van de $y$ -as.

c

${}^{2} log (x^{2}) = {}^{2} log (\frac{1}{2} x + 3)$ , dus $x^{2} = \frac{1}{2} x + 3$
$2 x^{2} - x - 6 = 0$ , dus $x = 2$ of $x = ‐ 1 \frac{1}{2}$ .

d

Omdat je de vergelijking ${}^{2} log (x^{2}) = {}^{2} log (\frac{1}{2} x + 3)$ oplost. En alleen voor positieve $x$ geldt: $2 \cdot log (x) = log (x^{2})$ !

3

$2 log (x - 2) = 1 + log (\frac{1}{2} x + 4)$
$log {(x - 2)}^{2} = log (10) + log (\frac{1}{2} x + 4)$
${(x - 2)}^{2} = 10 (\frac{1}{2} x + 4)$
$x^{2} - 9 x - 36 = 0$
$x = 12$ of $x = ‐ 3$
Alleen $x = 12$ is oplossing.
$2 log (x + 1) = 2 + log (x - 8)$
$log {(x + 1)}^{2} = log (100 (x - 8))$
${(x + 1)}^{2} - 100 (x - 8) = 0$
$x = 9$ of $x = 89$ , beide voldoen.
$log (x + 2) - log (x - 7) = log (2 x - 6)$
$log (\frac{x + 2}{x - 7}) = log (2 x - 6)$
$\frac{x + 2}{x - 7} = 2 x - 6$
$2 x^{2} - 21 x + 40 = 0$
$x = 2 \frac{1}{2}$ of $x = 8$
Alleen $x = 8$ is oplossing.

4

a

Invullen geeft $25$ km/u.

b

Als $s$ groter wordt, wordt de ${}^{2} log (...)$ groter en dus $20 - {}^{2} log (...)$ kleiner.

c

Als je $2 s$ in de formule $v = 20 - {}^{2} log (\frac{s}{10.000})$ invult in plaats van $s$ , krijg je: $v = 20 - {}^{2} log (\frac{2 s}{10.000}) = 20 - ({}^{2} log (2) + {}^{2} log (\frac{s}{10.000}))$
$= 20 - {}^{2} log (\frac{s}{10.000}) - 1$ , dus $v$ wordt $1$ kleiner want ${}^{2} log (2) = 1$ .

d

$v = 20 - {}^{2} log (\frac{s}{10.000})$
$v = {}^{2} log (2^{20}) - {}^{2} log (\frac{s}{10.000}) = {}^{2} log (\frac{10.000}{s}) \cdot 2^{20}$ , dus:
$2^{v} = \frac{2^{20} \cdot 10.000}{s}$ , dus $2^{v} \cdot s = 2^{20} \cdot 10.000 \approx 10^{10}$ , want $2^{10} \approx 1000 = 10^{3}$ .

5

a

Bij de gehoorgrens hoort geluidsdrukniveau $L = 10 \cdot log (\frac{I_{0}}{I_{0}}) = 0$ , want $log (1) = 0$ .
Bij de pijngrens hoort geluidsdrukniveau $10 \cdot log (\frac{10}{10^{‐ 12}}) = 10 \cdot log (10^{13}) = 130$ .

b

$10 \cdot log (\frac{2 I}{I_{0}}) = 10 \cdot log (2) + 10 \cdot log (\frac{I}{I_{0}}) = 10 \cdot log (2) + 80 = 83,0$

c

Noem de gezochte afstand $x$ , dan:
$74 = L_{0} - 10 log (2 π x)$ en
$77 = L_{0} - 10 log (40 π)$ , dus (trek de twee vergelijkingen van elkaar af en deel door $10$ )
$log (2 π x) = log (40 π) + 0,3 = log (40 π) \cdot 10^{0,3}$ , dus $x = 20 \cdot 10^{0,3} \approx 40$ meter.

6

$log (\frac{1}{5})$ $=$ $log (2) - log (10) = p - 1$	$log (3200)$ $=$ $log (32) + log (100) = 5 p + 2$
${}^{2} log (10)$ $=$ $\frac{log (10)}{log (2)} = \frac{1}{p}$	${}^{5} log (10)$ $=$ $\frac{log (10)}{log (5)} = \frac{log (10)}{log (10) - log (2)} = \frac{1}{1 - p}$