Formule bij exponentiële groei opstellen

\(y = b ⋅ g^{x}\) met \(g_{\text{week}} = 1 + {4{,}3 \over 100} = 1{,}043 \text{.}\)

De beginwaarde is de hoeveelheid bij \(x = 0 \text{,}\) dus \(b = 274 \text{.}\)

\(y = 274 ⋅ 1{,}043^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({412 \over 525})^{{1 \over 7 - 2}} = 0{,}952...\)

\(\begin{rcases}y = b ⋅ 0{,}952...^{x} \\ x = 2 \text{ en } y = 525\end{rcases} \begin{matrix}b ⋅ 0{,}952...^{2} = 525 \\ b = {525 \over 0{,}952...^{2}} ≈ 578\end{matrix}\)

\(y = 578 ⋅ 0{,}953^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({687 \over 594})^{{1 \over 6 - 2}} = 1{,}037...\)

\(\begin{rcases}y = b ⋅ 1{,}037...^{x} \\ x = 2 \text{ en } y = 594\end{rcases} \begin{matrix}b ⋅ 1{,}037...^{2} = 594 \\ b = {594 \over 1{,}037...^{2}} ≈ 552\end{matrix}\)

\(y = 552 ⋅ 1{,}037^{x} \text{.}\)

Rasterpunten \((2 , 90\,000)\) en \((8 , 40)\) aflezen.

\(y = b ⋅ g^{x}\) met \(g = ({40 \over 90\,000})^{{1 \over 8 - 2}} = 0{,}276...\)

\(\begin{rcases}y = b ⋅ 0{,}276...^{x} \\ x = 2 \text{ en } y = 90\,000{,}00\end{rcases} \begin{matrix}b ⋅ 0{,}276...^{2} = 90\,000{,}00 \\ b = {90\,000{,}00 \over 0{,}276...^{2}} \\ b = 1179333{,}627...\end{matrix}\)

\(y = 1\,179\,334 ⋅ 0{,}276^{x} \text{.}\)

\({30{,}44 \over 38{,}05} ≈ 0{,}80\)

\({24{,}35 \over 30{,}44} ≈ 0{,}80\)
\({19{,}48 \over 24{,}35} ≈ 0{,}80\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}8\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 38{,}05 \text{.}\)

Dus \(y = 38{,}05 ⋅ 0{,}80^{x} \text{.}\)

\(g = ({29{,}38 \over 18{,}80})^{{1 \over 2\,008 - 2\,006}} ≈ 1{,}25\)

\(g = ({71{,}72 \over 29{,}38})^{{1 \over 2\,012 - 2\,008}} ≈ 1{,}25\)
\(g = ({218{,}86 \over 71{,}72})^{{1 \over 2\,017 - 2\,012}} ≈ 1{,}25\)
\(g = ({834{,}89 \over 218{,}86})^{{1 \over 2\,023 - 2\,017}} ≈ 1{,}25\)

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

\(y = b ⋅ g^{x}\) met \(g = 1{,}25\)

\(\begin{rcases}y = b ⋅ 1{,}25^{x} \\ x = 1 \text{ en } y = 18{,}80\end{rcases} \begin{matrix}b ⋅ 1{,}25^{1} = 18{,}80 \\ b = {18{,}80 \over 1{,}25^{1}} \\ b ≈ 15{,}04\end{matrix}\)

Dus \(y = 15{,}04 ⋅ 1{,}25^{x} \text{.}\)

\(y = a x + b\) met \(a = {\Delta y \over \Delta x} = {4 - 9 \over 8 - 4} = -1{,}25\)

\(\begin{rcases}y = -1{,}25 x + b \\ \text{door } (4 , 9)\end{rcases} \begin{matrix}-1{,}25 ⋅ 4 + b = 9 \\ -5 + b = 9 \\ b = 14\end{matrix}\)

Dus \(l{:}\,y = -1{,}25 x + 14\)

\(y = b ⋅ g^{x}\) met \(g = ({4 \over 9})^{{1 \over 8 - 4}} = 0{,}816...\)

\(\begin{rcases}y = b ⋅ 0{,}816...^{x} \\ \text{door } (4 , 9)\end{rcases} \begin{matrix}9 = b ⋅ 0{,}816...^{4} \\ b = {9 \over 0{,}816...^{4}} \\ b = 20{,}25\end{matrix}\)

Dus \(y = 20{,}25 ⋅ 0{,}816^{x} \text{.}\)

\({12{,}99 \over 12{,}37} ≈ 1{,}05\)

\({13{,}64 \over 12{,}99} ≈ 1{,}05\)
\({14{,}32 \over 13{,}64} ≈ 1{,}05\)
\({15{,}04 \over 14{,}32} ≈ 1{,}05\)

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

\(y = b ⋅ g^{x}\) met \(g = 1{,}05\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 12{,}37 \text{.}\)

Dus \(y = 12{,}37 ⋅ 1{,}05^{x} \text{.}\)

\(g = ({93{,}35 \over 42{,}14})^{{1 \over 2\,015 - 2\,011}} ≈ 1{,}22\)

\(g = ({113{,}89 \over 93{,}35})^{{1 \over 2\,016 - 2\,015}} ≈ 1{,}22\)
\(g = ({307{,}81 \over 113{,}89})^{{1 \over 2\,021 - 2\,016}} ≈ 1{,}22\)
\(g = ({458{,}15 \over 307{,}81})^{{1 \over 2\,023 - 2\,021}} ≈ 1{,}22\)

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

\(y = b ⋅ g^{x}\) met \(g = 1{,}22\)

\(\begin{rcases}y = b ⋅ 1{,}22^{x} \\ x = 6 \text{ en } y = 42{,}14\end{rcases} \begin{matrix}b ⋅ 1{,}22^{6} = 42{,}14 \\ b = {42{,}14 \over 1{,}22^{6}} \\ b ≈ 12{,}78\end{matrix}\)

Dus \(y = 12{,}78 ⋅ 1{,}22^{x} \text{.}\)

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

Punt \(\text{B} (4 , 100\,000) \text{.}\)

Punt \(\text{C} (7 , 2\,000\,000) \text{.}\)

Rasterpunten \((20 , 3)\) en \((80 , 4)\) aflezen.

\(\log(y) = a x + b\) met \(a = {\Delta \log(y) \over \Delta x} = {4 - 3 \over 80 - 20} = \frac{1}{60}\)

\(\begin{rcases}\log(y) = \frac{1}{60} x + b \\ \text{door } (20 , 3)\end{rcases} \begin{matrix}\frac{1}{60} ⋅ 20 + b = 3 \\ \frac{20}{60} + b = 3 \\ b = 2\frac{2}{3}\end{matrix}\)

\(\log(y) = \frac{1}{60} x + 2\frac{2}{3}\)
\(y = 10^{\frac{1}{60} x + 2\frac{2}{3}}\)

\(y = 10^{\frac{1}{60} x} ⋅ 10^{2\frac{2}{3}}\)
\(\text{ } = 10^{2\frac{2}{3}} ⋅ (10^{\frac{1}{60}})^{x}\)
\(\text{ } = 464 ⋅ 1{,}039^{x}\)

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 \(y = a ⋅ \log(x) + b \text{.}\)