Formule bij exponentiële groei opstellen

\(y = b ⋅ g^{x}\) met \(g_{\text{uur}} = 1 + {2{,}1 \over 100} = 1{,}021 \text{.}\)

De beginwaarde is de hoeveelheid bij \(x = 0 \text{,}\) dus \(b = 427 \text{.}\)

\(y = 427 ⋅ 1{,}021^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({154 \over 197})^{{1 \over 9 - 4}} = 0{,}951...\)

\(\begin{rcases}y = b ⋅ 0{,}951...^{x} \\ x = 4 \text{ en } y = 197\end{rcases} \begin{matrix}b ⋅ 0{,}951...^{4} = 197 \\ b = {197 \over 0{,}951...^{4}} ≈ 240\end{matrix}\)

\(y = 240 ⋅ 0{,}952^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({551 \over 496})^{{1 \over 7 - 4}} = 1{,}035...\)

\(\begin{rcases}y = b ⋅ 1{,}035...^{x} \\ x = 4 \text{ en } y = 496\end{rcases} \begin{matrix}b ⋅ 1{,}035...^{4} = 496 \\ b = {496 \over 1{,}035...^{4}} ≈ 431\end{matrix}\)

\(y = 431 ⋅ 1{,}036^{x} \text{.}\)

Rasterpunten \((1 , 6\,000)\) en \((7 , 50)\) aflezen.

\(y = b ⋅ g^{x}\) met \(g = ({50 \over 6\,000})^{{1 \over 7 - 1}} = 0{,}450...\)

\(\begin{rcases}y = b ⋅ 0{,}450...^{x} \\ x = 1 \text{ en } y = 6\,000{,}00\end{rcases} \begin{matrix}b ⋅ 0{,}450...^{1} = 6\,000{,}00 \\ b = {6\,000{,}00 \over 0{,}450...^{1}} \\ b = 13325{,}436...\end{matrix}\)

\(y = 13\,325 ⋅ 0{,}450^{x} \text{.}\)

\({22{,}42 \over 18{,}68} ≈ 1{,}20\)

\({26{,}90 \over 22{,}42} ≈ 1{,}20\)
\({32{,}28 \over 26{,}90} ≈ 1{,}20\)
\({38{,}73 \over 32{,}28} ≈ 1{,}20\)

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

\(y = b ⋅ g^{x}\) met \(g = 1{,}2\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 18{,}68 \text{.}\)

Dus \(y = 18{,}68 ⋅ 1{,}20^{x} \text{.}\)

\(g = ({31{,}22 \over 45{,}26})^{{1 \over 7 - 1}} ≈ 0{,}94\)

\(g = ({27{,}59 \over 31{,}22})^{{1 \over 9 - 7}} ≈ 0{,}94\)
\(g = ({21{,}54 \over 27{,}59})^{{1 \over 13 - 9}} ≈ 0{,}94\)
\(g = ({15{,}81 \over 21{,}54})^{{1 \over 18 - 13}} ≈ 0{,}94\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}94\)

\(\begin{rcases}y = b ⋅ 0{,}94^{x} \\ x = 1 \text{ en } y = 45{,}26\end{rcases} \begin{matrix}b ⋅ 0{,}94^{1} = 45{,}26 \\ b = {45{,}26 \over 0{,}94^{1}} \\ b ≈ 48{,}15\end{matrix}\)

Dus \(y = 48{,}15 ⋅ 0{,}94^{x} \text{.}\)

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

\(\begin{rcases}y = 2 x + b \\ \text{door } (2 , 6)\end{rcases} \begin{matrix}2 ⋅ 2 + b = 6 \\ 4 + b = 6 \\ b = 2\end{matrix}\)

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

\(y = b ⋅ g^{x}\) met \(g = ({8 \over 6})^{{1 \over 3 - 2}} = 1{,}333...\)

\(\begin{rcases}y = b ⋅ 1{,}333...^{x} \\ \text{door } (2 , 6)\end{rcases} \begin{matrix}6 = b ⋅ 1{,}333...^{2} \\ b = {6 \over 1{,}333...^{2}} \\ b = 3{,}375\end{matrix}\)

Dus \(y = 3{,}375 ⋅ 1{,}333^{x} \text{.}\)

\({12{,}94 \over 10{,}78} ≈ 1{,}20\)

\({15{,}52 \over 12{,}94} ≈ 1{,}20\)
\({18{,}63 \over 15{,}52} ≈ 1{,}20\)

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

\(y = b ⋅ g^{x}\) met \(g = 1{,}2\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 10{,}78 \text{.}\)

Dus \(y = 10{,}78 ⋅ 1{,}20^{x} \text{.}\)

\({\Delta y \over \Delta x} = {10{,}37 - 10{,}43 \over 2\,021 - 2\,015} = -0{,}01\)

\({\Delta y \over \Delta x} = {10{,}36 - 10{,}37 \over 2\,022 - 2\,021} = -0{,}01\)
\({\Delta y \over \Delta x} = {10{,}34 - 10{,}36 \over 2\,024 - 2\,022} = -0{,}01\)

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

\(y = a x + b\) met \(a = -0{,}01\)

\(\begin{rcases}y = -0{,}01 x + b \\ x = 3 \text{ en } y = 10{,}43\end{rcases} \begin{matrix}-0{,}01 ⋅ 3 + b = 10{,}43 \\ -0{,}03 + b = 10{,}43 \\ b = 10{,}46\end{matrix}\)

Dus \(y = -0{,}01 x + 10{,}46\)

Punt \(\text{A} (2 , 50) \text{.}\)

Punt \(\text{B} (4 , 800) \text{.}\)

Punt \(\text{C} (7 , 1\,000) \text{.}\)

Rasterpunten \((2 , 3)\) en \((12 , 6)\) aflezen.

\(\log(y) = a x + b\) met \(a = {\Delta \log(y) \over \Delta x} = {6 - 3 \over 12 - 2} = \frac{3}{10}\)

\(\begin{rcases}\log(y) = \frac{3}{10} x + b \\ \text{door } (2 , 3)\end{rcases} \begin{matrix}\frac{3}{10} ⋅ 2 + b = 3 \\ \frac{6}{10} + b = 3 \\ b = 2\frac{2}{5}\end{matrix}\)

\(\log(y) = \frac{3}{10} x + 2\frac{2}{5}\)
\(y = 10^{\frac{3}{10} x + 2\frac{2}{5}}\)

\(y = 10^{\frac{3}{10} x} ⋅ 10^{2\frac{2}{5}}\)
\(\text{ } = 10^{2\frac{2}{5}} ⋅ (10^{\frac{3}{10}})^{x}\)
\(\text{ } = 251 ⋅ 1{,}995^{x}\)

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 lineair verband.

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