Formule bij exponentiële groei opstellen

\(y = b ⋅ g^{x}\) met \(g_{\text{jaar}} = 1 - {2{,}2 \over 100} = 0{,}978 \text{.}\)

De beginwaarde is de hoeveelheid bij \(x = 0 \text{,}\) dus \(b = 279 \text{.}\)

\(y = 279 ⋅ 0{,}978^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({483 \over 552})^{{1 \over 6 - 2}} = 0{,}967...\)

\(\begin{rcases}y = b ⋅ 0{,}967...^{x} \\ x = 2 \text{ en } y = 552\end{rcases} \begin{matrix}b ⋅ 0{,}967...^{2} = 552 \\ b = {552 \over 0{,}967...^{2}} ≈ 590\end{matrix}\)

\(y = 590 ⋅ 0{,}967^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({280 \over 259})^{{1 \over 5 - 3}} = 1{,}039...\)

\(\begin{rcases}y = b ⋅ 1{,}039...^{x} \\ x = 3 \text{ en } y = 259\end{rcases} \begin{matrix}b ⋅ 1{,}039...^{3} = 259 \\ b = {259 \over 1{,}039...^{3}} ≈ 230\end{matrix}\)

\(y = 230 ⋅ 1{,}040^{x} \text{.}\)

Rasterpunten \((2 , 7\,000)\) en \((7 , 300)\) aflezen.

\(y = b ⋅ g^{x}\) met \(g = ({300 \over 7\,000})^{{1 \over 7 - 2}} = 0{,}532...\)

\(\begin{rcases}y = b ⋅ 0{,}532...^{x} \\ x = 2 \text{ en } y = 7\,000{,}00\end{rcases} \begin{matrix}b ⋅ 0{,}532...^{2} = 7\,000{,}00 \\ b = {7\,000{,}00 \over 0{,}532...^{2}} \\ b = 24676{,}795...\end{matrix}\)

\(y = 24\,677 ⋅ 0{,}533^{x} \text{.}\)

\({24{,}84 \over 33{,}12} ≈ 0{,}75\)

\({18{,}63 \over 24{,}84} ≈ 0{,}75\)
\({13{,}97 \over 18{,}63} ≈ 0{,}75\)
\({10{,}48 \over 13{,}97} ≈ 0{,}75\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}75\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 33{,}12 \text{.}\)

Dus \(y = 33{,}12 ⋅ 0{,}75^{x} \text{.}\)

\(g = ({142{,}92 \over 289{,}87})^{{1 \over 2\,012 - 2\,009}} ≈ 0{,}79\)

\(g = ({43{,}98 \over 142{,}92})^{{1 \over 2\,017 - 2\,012}} ≈ 0{,}79\)
\(g = ({10{,}69 \over 43{,}98})^{{1 \over 2\,023 - 2\,017}} ≈ 0{,}79\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}79\)

\(\begin{rcases}y = b ⋅ 0{,}79^{x} \\ x = 4 \text{ en } y = 289{,}87\end{rcases} \begin{matrix}b ⋅ 0{,}79^{4} = 289{,}87 \\ b = {289{,}87 \over 0{,}79^{4}} \\ b ≈ 744{,}21\end{matrix}\)

Dus \(y = 744{,}21 ⋅ 0{,}79^{x} \text{.}\)

\(y = a x + b\) met \(a = {\Delta y \over \Delta x} = {4 - 5 \over 10 - 5} = -0{,}2\)

\(\begin{rcases}y = -0{,}2 x + b \\ \text{door } (5 , 5)\end{rcases} \begin{matrix}-0{,}2 ⋅ 5 + b = 5 \\ -1 + b = 5 \\ b = 6\end{matrix}\)

Dus \(l{:}\,y = -0{,}2 x + 6\)

\(y = b ⋅ g^{x}\) met \(g = ({4 \over 5})^{{1 \over 10 - 5}} = 0{,}956...\)

\(\begin{rcases}y = b ⋅ 0{,}956...^{x} \\ \text{door } (5 , 5)\end{rcases} \begin{matrix}5 = b ⋅ 0{,}956...^{5} \\ b = {5 \over 0{,}956...^{5}} \\ b = 6{,}25\end{matrix}\)

Dus \(y = 6{,}25 ⋅ 0{,}956^{x} \text{.}\)

\(16{,}16 - 15{,}67 = 0{,}49\)

\(16{,}65 - 16{,}16 = 0{,}49\)
\(17{,}14 - 16{,}65 = 0{,}49\)
\(17{,}63 - 17{,}14 = 0{,}49\)

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

\(y = a x + b\) met \(a = 0{,}49\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 15{,}67 \text{.}\)

Dus \(y = 0{,}49 x + 15{,}67\)

\(g = ({21{,}41 \over 32{,}64})^{{1 \over 10 - 6}} ≈ 0{,}90\)

\(g = ({17{,}34 \over 21{,}41})^{{1 \over 12 - 10}} ≈ 0{,}90\)
\(g = ({12{,}64 \over 17{,}34})^{{1 \over 15 - 12}} ≈ 0{,}90\)
\(g = ({11{,}38 \over 12{,}64})^{{1 \over 16 - 15}} ≈ 0{,}90\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}9\)

\(\begin{rcases}y = b ⋅ 0{,}9^{x} \\ x = 6 \text{ en } y = 32{,}64\end{rcases} \begin{matrix}b ⋅ 0{,}9^{6} = 32{,}64 \\ b = {32{,}64 \over 0{,}9^{6}} \\ b ≈ 61{,}42\end{matrix}\)

Dus \(y = 61{,}42 ⋅ 0{,}90^{x} \text{.}\)

Punt \(\text{A} (1 , 20\,000) \text{.}\)

Punt \(\text{B} (5 , 500\,000) \text{.}\)

Punt \(\text{C} (7 , 100) \text{.}\)

Rasterpunten \((2 , 3)\) en \((8 , 2)\) aflezen.

\(\log(y) = a x + b\) met \(a = {\Delta \log(y) \over \Delta x} = {2 - 3 \over 8 - 2} = -\frac{1}{6}\)

\(\begin{rcases}\log(y) = -\frac{1}{6} x + b \\ \text{door } (2 , 3)\end{rcases} \begin{matrix}-\frac{1}{6} ⋅ 2 + b = 3 \\ -\frac{2}{6} + b = 3 \\ b = 3\frac{1}{3}\end{matrix}\)

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

\(y = 10^{-\frac{1}{6} x} ⋅ 10^{3\frac{1}{3}}\)
\(\text{ } = 10^{3\frac{1}{3}} ⋅ (10^{-\frac{1}{6}})^{x}\)
\(\text{ } = 2\,154 ⋅ 0{,}681^{x}\)

Rasterpunten \((2 , 4)\) en \((5 , 6)\) aflezen.

\(\log(y) = a ⋅ \log(x) + b\) met \(a = {\Delta \log(y) \over \Delta \log(x)} = {6 - 4 \over 5 - 2} = \frac{2}{3}\)

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

\(\log(y) = \frac{2}{3} ⋅ \log(x) + 2\frac{2}{3}\)
\(\log(y) = \log(x^{\frac{2}{3}}) + \log(10^{2\frac{2}{3}})\)
\(\log(y) = \log(10^{2\frac{2}{3}} ⋅ x^{\frac{2}{3}})\)

\(y = 10^{2\frac{2}{3}} ⋅ x^{\frac{2}{3}}\)
\(y = 464 ⋅ x^{0{,}67}\)

De grafiek hoort bij een logaritmisch verband.

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