Formule bij exponentiële groei opstellen

\(y = b ⋅ g^{x}\) met \(g_{\text{uur}} = 1 - {1{,}1 \over 100} = 0{,}989 \text{.}\)

De beginwaarde is de hoeveelheid bij \(x = 0 \text{,}\) dus \(b = 533 \text{.}\)

\(y = 533 ⋅ 0{,}989^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({225 \over 286})^{{1 \over 9 - 4}} = 0{,}953...\)

\(\begin{rcases}y = b ⋅ 0{,}953...^{x} \\ x = 4 \text{ en } y = 286\end{rcases} \begin{matrix}b ⋅ 0{,}953...^{4} = 286 \\ b = {286 \over 0{,}953...^{4}} ≈ 347\end{matrix}\)

\(y = 347 ⋅ 0{,}953^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({795 \over 657})^{{1 \over 10 - 5}} = 1{,}038...\)

\(\begin{rcases}y = b ⋅ 1{,}038...^{x} \\ x = 5 \text{ en } y = 657\end{rcases} \begin{matrix}b ⋅ 1{,}038...^{5} = 657 \\ b = {657 \over 1{,}038...^{5}} ≈ 543\end{matrix}\)

\(y = 543 ⋅ 1{,}039^{x} \text{.}\)

Rasterpunten \((2 , 50)\) en \((8 , 30\,000)\) aflezen.

\(y = b ⋅ g^{x}\) met \(g = ({30\,000 \over 50})^{{1 \over 8 - 2}} = 2{,}904...\)

\(\begin{rcases}y = b ⋅ 2{,}904...^{x} \\ x = 2 \text{ en } y = 50{,}00\end{rcases} \begin{matrix}b ⋅ 2{,}904...^{2} = 50{,}00 \\ b = {50{,}00 \over 2{,}904...^{2}} \\ b = 5{,}928...\end{matrix}\)

\(y = 6 ⋅ 2{,}904^{x} \text{.}\)

\({50{,}71 \over 71{,}42} ≈ 0{,}71\)

\({36{,}00 \over 50{,}71} ≈ 0{,}71\)
\({25{,}56 \over 36{,}00} ≈ 0{,}71\)
\({18{,}15 \over 25{,}56} ≈ 0{,}71\)

De quotiënten zijn bij benadering gelijk, dus de tabel hoort bij een exponentieel verband.

\(y = b ⋅ g^{x}\) met \(g = 0{,}71\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 71{,}42 \text{.}\)

Dus \(y = 71{,}42 ⋅ 0{,}71^{x} \text{.}\)

\(g = ({13{,}27 \over 13{,}96})^{{1 \over 7 - 2}} ≈ 0{,}99\)

\(g = ({12{,}75 \over 13{,}27})^{{1 \over 11 - 7}} ≈ 0{,}99\)
\(g = ({12{,}37 \over 12{,}75})^{{1 \over 14 - 11}} ≈ 0{,}99\)
\(g = ({12{,}25 \over 12{,}37})^{{1 \over 15 - 14}} ≈ 0{,}99\)

De groeifactoren zijn bij benadering gelijk, dus de tabel hoort bij een exponentieel verband.

\(y = b ⋅ g^{x}\) met \(g = 0{,}99\)

\(\begin{rcases}y = b ⋅ 0{,}99^{x} \\ x = 2 \text{ en } y = 13{,}96\end{rcases} \begin{matrix}b ⋅ 0{,}99^{2} = 13{,}96 \\ b = {13{,}96 \over 0{,}99^{2}} \\ b ≈ 14{,}24\end{matrix}\)

Dus \(y = 14{,}24 ⋅ 0{,}99^{x} \text{.}\)

\(y = a x + b\) met \(a = {\Delta y \over \Delta x} = {5 - 9 \over 6 - 4} = -2\)

\(\begin{rcases}y = -2 x + b \\ \text{door } (4 , 9)\end{rcases} \begin{matrix}-2 ⋅ 4 + b = 9 \\ -8 + b = 9 \\ b = 17\end{matrix}\)

Dus \(l{:}\,y = -2 x + 17\)

\(y = b ⋅ g^{x}\) met \(g = ({5 \over 9})^{{1 \over 6 - 4}} = 0{,}745...\)

\(\begin{rcases}y = b ⋅ 0{,}745...^{x} \\ \text{door } (4 , 9)\end{rcases} \begin{matrix}9 = b ⋅ 0{,}745...^{4} \\ b = {9 \over 0{,}745...^{4}} \\ b = 29{,}16\end{matrix}\)

Dus \(y = 29{,}16 ⋅ 0{,}745^{x} \text{.}\)

\(12{,}41 - 11{,}68 = 0{,}73\)

\(13{,}14 - 12{,}41 = 0{,}73\)
\(13{,}87 - 13{,}14 = 0{,}73\)
\(14{,}60 - 13{,}87 = 0{,}73\)

Het verschil is steeds hetzelfde, dus de tabel hoort bij een lineair verband.

\(y = a x + b\) met \(a = 0{,}73\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 11{,}68 \text{.}\)

Dus \(y = 0{,}73 x + 11{,}68\)

\({\Delta y \over \Delta x} = {37{,}01 - 25{,}79 \over 11 - 5} = 1{,}87\)

\({\Delta y \over \Delta x} = {42{,}62 - 37{,}01 \over 14 - 11} = 1{,}87\)
\({\Delta y \over \Delta x} = {46{,}36 - 42{,}62 \over 16 - 14} = 1{,}87\)

De gemiddelde verandering is steeds hetzelfde, dus de tabel hoort bij een lineair verband.

\(y = a x + b\) met \(a = 1{,}87\)

\(\begin{rcases}y = 1{,}87 x + b \\ x = 5 \text{ en } y = 25{,}79\end{rcases} \begin{matrix}1{,}87 ⋅ 5 + b = 25{,}79 \\ 9{,}35 + b = 25{,}79 \\ b = 16{,}44\end{matrix}\)

Dus \(y = 1{,}87 x + 16{,}44\)

Punt \(\text{A} (3 , 5\,000) \text{.}\)

Punt \(\text{B} (4 , 600) \text{.}\)

Punt \(\text{C} (9 , 10\,000) \text{.}\)

Rasterpunten \((1 , 4)\) en \((8 , 5)\) aflezen.

\(\log(y) = a x + b\) met \(a = {\Delta \log(y) \over \Delta x} = {5 - 4 \over 8 - 1} = \frac{1}{7}\)

\(\begin{rcases}\log(y) = \frac{1}{7} x + b \\ \text{door } (1 , 4)\end{rcases} \begin{matrix}\frac{1}{7} ⋅ 1 + b = 4 \\ \frac{1}{7} + b = 4 \\ b = 3\frac{6}{7}\end{matrix}\)

\(\log(y) = \frac{1}{7} x + 3\frac{6}{7}\)
\(y = 10^{\frac{1}{7} x + 3\frac{6}{7}}\)

\(y = 10^{\frac{1}{7} x} ⋅ 10^{3\frac{6}{7}}\)
\(\text{ } = 10^{3\frac{6}{7}} ⋅ (10^{\frac{1}{7}})^{x}\)
\(\text{ } = 7\,197 ⋅ 1{,}389^{x}\)

Rasterpunten \((1 , 2)\) en \((5 , 3)\) aflezen.

\(\log(y) = a ⋅ \log(x) + b\) met \(a = {\Delta \log(y) \over \Delta \log(x)} = {3 - 2 \over 5 - 1} = \frac{1}{4}\)

\(\begin{rcases}\log(y) = \frac{1}{4} ⋅ \log(x) + b \\ \text{door } (1 , 2)\end{rcases} \begin{matrix}\frac{1}{4} ⋅ 1 + b = 2 \\ \frac{1}{4} + b = 2 \\ b = 1\frac{3}{4}\end{matrix}\)

\(\log(y) = \frac{1}{4} ⋅ \log(x) + 1\frac{3}{4}\)
\(\log(y) = \log(x^{\frac{1}{4}}) + \log(10^{1\frac{3}{4}})\)
\(\log(y) = \log(10^{1\frac{3}{4}} ⋅ x^{\frac{1}{4}})\)

\(y = 10^{1\frac{3}{4}} ⋅ x^{\frac{1}{4}}\)
\(y = 56 ⋅ x^{0{,}25}\)

De grafiek hoort bij een logaritmisch verband.

Hierbij hoort de formule \(y = a ⋅ \log(x) + b \text{.}\)