$t=0$: $\log (L)=\mathrm{0,9717}$ $\to$ $L={10}^{\mathrm{0,9717}}\approx \mathrm{9,369}$;
$t=85$: $\log (L)=\mathrm{1,7537}$ $\to$ $L={10}^{\mathrm{1,7537}}\approx \mathrm{56,715}$;
De berekende waarden komen goed overeen met de gemeten waarden.

$\log (75)\approx \mathrm{1,875...}$, dus $\mathrm{1,875...}=\mathrm{0,0092}t+\mathrm{0,9717}$ $\to$ $t\approx \mathrm{98,2}$, dus na $98$ dagen.

$L={10}^{\mathrm{0,0092}t+\mathrm{0,9717}}={10}^{\mathrm{0,9717}}\cdot {\left({10}^{\mathrm{0,0092}}\right)}^t\approx \mathrm{9,369}\cdot {\mathrm{1,0214}}^t$

$g={10}^{\mathrm{0,0092}}\approx \mathrm{1,021}$, dus de groei is $\mathrm{2,1}\text{\%}$ per dag

$\log (\mathrm{108,2})-\log (\mathrm{57,9})=a\cdot (\log (225)-\log (88))$ $\to$ $\mathrm{0,2715...}=a\cdot \mathrm{0,4076...}$ $\to$ $a=\frac{\mathrm{0,2715...}}{\mathrm{0,4076...}}\approx \mathrm{0,666}$
Invullen geeft: $b=\log (\mathrm{57,9})-\mathrm{0,666...}\times \log (88)\approx \mathrm{0,468}$

Klopt.

$\log (R)=\log (T^{\mathrm{0,667}})+\log ({10}^{\mathrm{0,466}})=\log ({10}^{\mathrm{0,466}}\cdot T^{\mathrm{0,667}})\approx \log (\mathrm{2,924}\cdot T^{\mathrm{0,667}})$

$8^{\mathrm{0,667}}\approx 4$ keer zo groot.

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Een rechte lijn

• $\log (y)=\log (x^2)+\log (3)=\log (3x^2)$ $\to$ $y=3x^2$

• $\log (y)=\log ({10}^{1\frac{1}{2}})-\log (x)=\log (\frac{{10}^{1\frac{1}{2}}}{x})=\log (\frac{10\sqrt{10}}{x})$
  $\to$ $y=\frac{10\sqrt{10}}{x}$ (of: $y=10\sqrt{10}\cdot x^{\mathrm{‐}1}$)

• $\log (y)=\log (x^{\frac{1}{2}})+\log ({10}^3)$ $\to$ $\log (y)=\log ({10}^3\cdot x^{\frac{1}{2}})=\log (1000\sqrt{x})$
  $\to$ $y=1000\sqrt{x}$

• $\log (y)=\log (10\cdot x^4)=\log (10)+\log (x^4)=4\cdot \log (x)+1$

• $\log (y)=\log (\mathrm{0,2}\cdot \sqrt{x})=\log (\mathrm{0,2})+\log (x^{\frac{1}{2}})\approx \frac{1}{2}\cdot \log (x)-\mathrm{0,699}$

• $\log (y)=\log \left(\frac{8}{x}\right)=\log (8)-\log (x)\approx \mathrm{‐}\log (x)+\mathrm{0,903}$

$\log (T)=\mathrm{1,07}+\mathrm{0,20}\cdot \log (4000)\approx \mathrm{1,790...}$ $\to$ $T={10}^{\mathrm{1,790...}}\approx \mathrm{61,7}$ (of $62$) jaar

$\log (44)=\mathrm{1,07}+\mathrm{0,20}\cdot \log (G)$ $\to$ $\log (G)=\mathrm{2,867...}$ $\to$ $G={10}^{\mathrm{2,867...}}\approx \mathrm{736,7}$, dus $740$ kg.

richtingscoëfficiënt $=\mathrm{0,20}$

$\log (T)=\mathrm{1,07}+\mathrm{0,20}\cdot \log (2G)=\mathrm{1,07}+\mathrm{0,20}\cdot \log (G)+\mathrm{0,20}\cdot \log (2)$, dus $\log (T)$ neemt toe met $\mathrm{0,20}\cdot \log (2)\approx \mathrm{0,0602}$; dan wordt $T$ vermenigvuldigd met ${10}^{\mathrm{0,0602...}}\approx \mathrm{1,15}$, ofwel neemt met $15\text{\%}$ toe

$\log (G)=\log ({10}^{\mathrm{1,07}})+\log (G^{\mathrm{0,20}})=\log ({10}^{\mathrm{1,07}}\cdot G^{\mathrm{0,20}})\approx \log (\mathrm{11,75}\cdot G^{\mathrm{0,20}})$, dus $G=\mathrm{11,75}\cdot G^{\mathrm{0,20}}$; $a=\mathrm{11,75}$ en $b=\mathrm{0,20}$

${10}^{\mathrm{‐}\mathrm{0,35}}\approx \mathrm{0,45}$ kg, dus $450$ gram

Aflezen: $\log (\frac{W}{40})=\mathrm{0,30}$, dus $\frac{W}{40}={10}^{\mathrm{0,30}}\approx \mathrm{1,995...}$; $W\approx 79$ (joules)

$\log (\frac{W}{40})=\log ({10}^{\mathrm{0,50}})+\log (G^{\mathrm{0,75}})=\log ({10}^{\mathrm{0,50}}\cdot G^{\mathrm{0,75}})$ $\to$ $\frac{W}{40}={10}^{\mathrm{0,50}}\cdot G^{\mathrm{0,75}}$ $\to$ $W=40\cdot {10}^{\mathrm{0,50}}\cdot G^{\mathrm{0,75}}\approx \mathrm{126,5}\cdot G^{\mathrm{0,75}}$

Bij de gehoorgrens hoort geluidsdrukniveau $L=10\cdot \text{log}(\frac{I_0}{I_0})=0$, want $\text{log}(1)=0$;
Bij de pijngrens hoort geluidsdrukniveau $10\cdot \text{log}(\frac{10}{{10}^{\mathrm{‐}12}})=10\cdot \text{log}({10}^{13})=130$

$10\cdot \text{log}(\frac{2I}{I_0})=10\cdot \text{log}(2)+10\cdot \text{log}(\frac{I}{I_0})=10\cdot \text{log}(2)+80=\mathrm{83,0}$

Noem de gezochte afstand $x$, dan:
$74=L_0-10\text{log}(2\text{π}x)$ en $77=L_0-10\text{log}(40\text{π})$,
dus (trek de twee vergelijkingen van elkaar af en deel door $10$)
$\mathrm{0,3}=\log (2\text{π}x)-\log (40\text{π})$ $\to$ $\text{log}(2\text{π}x)=\text{log}(40\text{π})+\mathrm{0,3}=\text{log}(40\text{π}\cdot {10}^{\mathrm{0,3}})$, dus $x=20\cdot {10}^{\mathrm{0,3}}\approx 40$ meter

• ${}^2\log (y)=\hspace{0.17em}{}^2\log (4\cdot 2^x)=\hspace{0.17em}{}^2\log (4)+{}^2\log (2^x)=\hspace{0.17em}{}^2\log (4)+x\cdot {}^2\log (2)=2+x$ $\to$ $x=\hspace{0.17em}{}^2\log (y)-2$

• $y-3=4\cdot 2^x$ $\to$ ${}^2\log (y-3)=\hspace{0.17em}{}^2\log (4\cdot 2^x)=\hspace{0.17em}{}^2\log (4)+{}^2\log (2^x)=\hspace{0.17em}{}^2\log (4)+x\cdot {}^2\log (2)=2+x$ $\to$ $x=\hspace{0.17em}{}^2\log (y-3)-2$

• ${}^3\log (y)=\hspace{0.17em}{}^3\log (\frac{1}{3}\cdot 3^x)=\hspace{0.17em}{}^3\log (\frac{1}{3})+{}^3\log (3^x)=\mathrm{‐}1+x\cdot {}^3\log (3)=\mathrm{‐}1+x$ $\to$ $x=\hspace{0.17em}{}^3\log (y)+1$

• $9\cdot 3^x=2-y$ $\to$ ${}^3\log (2-y)=\hspace{0.17em}{}^3\log (9\cdot 3^x)=\hspace{0.17em}{}^3\log (9)+{}^3\log (3^x)=2+x$ $\to$ $x=\hspace{0.17em}{}^3\log (2-y)-2$

$\log (L)=\log (\mathrm{9,37}\cdot {\mathrm{1,02}}^t)=\log (\mathrm{9,37})+t\cdot \log (\mathrm{1,02})$ $\to$ $t=\mathrm{‐}\frac{\log (\mathrm{9,37})}{\log (\mathrm{1,02})}+\frac{1}{\log (\mathrm{1,02})}\cdot \log (L)\approx \mathrm{‐}\mathrm{113,0}+\mathrm{116,3}\cdot \log (L)$ (dus $a\approx \mathrm{‐}\mathrm{113,0}$ en $b\approx \mathrm{116,3}$)

Fosfaatrijk: $\log (A)=\log (\mathrm{10,0}\cdot {\mathrm{1,6}}^t)=\log (\mathrm{10,0})+t\cdot \log (\mathrm{1,6})=1+t\cdot \log (\mathrm{1,6})$ $\to$ $t=\frac{1}{\log (\mathrm{1,6})}\cdot \log (A)-\frac{1}{\log (\mathrm{1,6})}\approx \mathrm{4,90}\cdot \log (A)-\mathrm{4,90}$
Fosfaatarm: $\log (A)=\log (\mathrm{10,0}\cdot {\mathrm{1,3}}^t)=\log (\mathrm{10,0})+t\cdot \log (\mathrm{1,3})=1+t\cdot \log (\mathrm{1,3})$ $\to$ $t=\frac{1}{\log (\mathrm{1,3})}\cdot \log (A)-\frac{1}{\log (\mathrm{1,3})}\approx \mathrm{8,78}\cdot \log (A)-\mathrm{8,78}$

$\frac{\mathrm{10,0}\cdot {\mathrm{1,6}}^t}{\mathrm{10,0}\cdot {\mathrm{1,3}}^t}=2$ $\to$ $\frac{{\mathrm{1,6}}^t}{{\mathrm{1,3}}^t}=2$ $\to$ ${\left(\frac{\mathrm{1,6}}{\mathrm{1,3}}\right)}^t=2$ $\to$ ${\mathrm{1,23...}}^t=2$ $\to$ $t=\hspace{0.17em}{}^{\mathrm{1,23...}}\log (2)\approx \mathrm{3,338}$ weken, dus na $\mathrm{23,4}$ dagen.