Formule bij exponentiële groei opstellen

\(y = b ⋅ g^{x}\) met \(g_{\text{week}} = 1 - {1{,}4 \over 100} = 0{,}986 \text{.}\)

De beginwaarde is de hoeveelheid bij \(x = 0 \text{,}\) dus \(b = 405 \text{.}\)

\(y = 405 ⋅ 0{,}986^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({387 \over 453})^{{1 \over 8 - 4}} = 0{,}961...\)

\(\begin{rcases}y = b ⋅ 0{,}961...^{x} \\ x = 4 \text{ en } y = 453\end{rcases} \begin{matrix}b ⋅ 0{,}961...^{4} = 453 \\ b = {453 \over 0{,}961...^{4}} ≈ 530\end{matrix}\)

\(y = 530 ⋅ 0{,}961^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({308 \over 283})^{{1 \over 4 - 2}} = 1{,}043...\)

\(\begin{rcases}y = b ⋅ 1{,}043...^{x} \\ x = 2 \text{ en } y = 283\end{rcases} \begin{matrix}b ⋅ 1{,}043...^{2} = 283 \\ b = {283 \over 1{,}043...^{2}} ≈ 260\end{matrix}\)

\(y = 260 ⋅ 1{,}043^{x} \text{.}\)

Rasterpunten \((1 , 200)\) en \((7 , 3\,000)\) aflezen.

\(y = b ⋅ g^{x}\) met \(g = ({3\,000 \over 200})^{{1 \over 7 - 1}} = 1{,}570...\)

\(\begin{rcases}y = b ⋅ 1{,}570...^{x} \\ x = 1 \text{ en } y = 200{,}00\end{rcases} \begin{matrix}b ⋅ 1{,}570...^{1} = 200{,}00 \\ b = {200{,}00 \over 1{,}570...^{1}} \\ b = 127{,}354...\end{matrix}\)

\(y = 127 ⋅ 1{,}570^{x} \text{.}\)

\({23{,}49 \over 33{,}08} ≈ 0{,}71\)

\({16{,}68 \over 23{,}49} ≈ 0{,}71\)
\({11{,}84 \over 16{,}68} ≈ 0{,}71\)

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 \(b = 33{,}08 \text{.}\)

Dus \(y = 33{,}08 ⋅ 0{,}71^{x} \text{.}\)

\(g = ({1\,095{,}52 \over 1\,947{,}60})^{{1 \over 5 - 3}} ≈ 0{,}75\)

\(g = ({346{,}63 \over 1\,095{,}52})^{{1 \over 9 - 5}} ≈ 0{,}75\)
\(g = ({61{,}69 \over 346{,}63})^{{1 \over 15 - 9}} ≈ 0{,}75\)
\(g = ({14{,}64 \over 61{,}69})^{{1 \over 20 - 15}} ≈ 0{,}75\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}75\)

\(\begin{rcases}y = b ⋅ 0{,}75^{x} \\ x = 3 \text{ en } y = 1\,947{,}60\end{rcases} \begin{matrix}b ⋅ 0{,}75^{3} = 1\,947{,}60 \\ b = {1\,947{,}60 \over 0{,}75^{3}} \\ b ≈ 4\,616{,}53\end{matrix}\)

Dus \(y = 4\,616{,}53 ⋅ 0{,}75^{x} \text{.}\)

\(y = a x + b\) met \(a = {\Delta y \over \Delta x} = {1 - 11 \over 12 - 11} = -10\)

\(\begin{rcases}y = -10 x + b \\ \text{door } (11 , 11)\end{rcases} \begin{matrix}-10 ⋅ 11 + b = 11 \\ -110 + b = 11 \\ b = 121\end{matrix}\)

Dus \(l{:}\,y = -10 x + 121\)

\(y = b ⋅ g^{x}\) met \(g = ({1 \over 11})^{{1 \over 12 - 11}} = 0{,}090...\)

\(\begin{rcases}y = b ⋅ 0{,}090...^{x} \\ \text{door } (11 , 11)\end{rcases} \begin{matrix}11 = b ⋅ 0{,}090...^{11} \\ b = {11 \over 0{,}090...^{11}} \\ b = 3\,138\,428\,376\,721\end{matrix}\)

Dus \(y = 3\,138\,428\,376\,721 ⋅ 0{,}091^{x} \text{.}\)

\(17{,}52 - 16{,}33 = 1{,}19\)

\(18{,}71 - 17{,}52 = 1{,}19\)
\(19{,}90 - 18{,}71 = 1{,}19\)
\(21{,}09 - 19{,}90 = 1{,}19\)

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

\(y = a x + b\) met \(a = 1{,}19\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 16{,}33 \text{.}\)

Dus \(y = 1{,}19 x + 16{,}33\)

\(g = ({153{,}67 \over 390{,}12})^{{1 \over 2\,014 - 2\,009}} ≈ 0{,}83\)

\(g = ({105{,}86 \over 153{,}67})^{{1 \over 2\,016 - 2\,014}} ≈ 0{,}83\)
\(g = ({60{,}53 \over 105{,}86})^{{1 \over 2\,019 - 2\,016}} ≈ 0{,}83\)
\(g = ({19{,}79 \over 60{,}53})^{{1 \over 2\,025 - 2\,019}} ≈ 0{,}83\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}83\)

\(\begin{rcases}y = b ⋅ 0{,}83^{x} \\ x = 4 \text{ en } y = 390{,}12\end{rcases} \begin{matrix}b ⋅ 0{,}83^{4} = 390{,}12 \\ b = {390{,}12 \over 0{,}83^{4}} \\ b ≈ 822{,}03\end{matrix}\)

Dus \(y = 822{,}03 ⋅ 0{,}83^{x} \text{.}\)

Punt \(\text{A} (2 , 40\,000) \text{.}\)

Punt \(\text{B} (5 , 1\,000) \text{.}\)

Punt \(\text{C} (8 , 900) \text{.}\)

Rasterpunten \((10 , 3)\) en \((35 , 6)\) aflezen.

\(\log(y) = a x + b\) met \(a = {\Delta \log(y) \over \Delta x} = {6 - 3 \over 35 - 10} = \frac{3}{25}\)

\(\begin{rcases}\log(y) = \frac{3}{25} x + b \\ \text{door } (10 , 3)\end{rcases} \begin{matrix}\frac{3}{25} ⋅ 10 + b = 3 \\ 1\frac{5}{25} + b = 3 \\ b = 1\frac{4}{5}\end{matrix}\)

\(\log(y) = \frac{3}{25} x + 1\frac{4}{5}\)
\(y = 10^{\frac{3}{25} x + 1\frac{4}{5}}\)

\(y = 10^{\frac{3}{25} x} ⋅ 10^{1\frac{4}{5}}\)
\(\text{ } = 10^{1\frac{4}{5}} ⋅ (10^{\frac{3}{25}})^{x}\)
\(\text{ } = 63 ⋅ 1{,}318^{x}\)

Rasterpunten \((1 , 4)\) en \((6 , 6)\) aflezen.

\(\log(y) = a ⋅ \log(x) + b\) met \(a = {\Delta \log(y) \over \Delta \log(x)} = {6 - 4 \over 6 - 1} = \frac{2}{5}\)

\(\begin{rcases}\log(y) = \frac{2}{5} ⋅ \log(x) + b \\ \text{door } (1 , 4)\end{rcases} \begin{matrix}\frac{2}{5} ⋅ 1 + b = 4 \\ \frac{2}{5} + b = 4 \\ b = 3\frac{3}{5}\end{matrix}\)

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

\(y = 10^{3\frac{3}{5}} ⋅ x^{\frac{2}{5}}\)
\(y = 3\,981 ⋅ x^{0{,}40}\)

De grafiek hoort bij een logaritmisch verband.

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