Formule bij exponentiële groei opstellen

\(y = b ⋅ g^{x}\) met \(g_{\text{dag}} = 1 - {2{,}7 \over 100} = 0{,}973 \text{.}\)

De beginwaarde is de hoeveelheid bij \(x = 0 \text{,}\) dus \(b = 277 \text{.}\)

\(y = 277 ⋅ 0{,}973^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({218 \over 238})^{{1 \over 6 - 4}} = 0{,}957...\)

\(\begin{rcases}y = b ⋅ 0{,}957...^{x} \\ x = 4 \text{ en } y = 238\end{rcases} \begin{matrix}b ⋅ 0{,}957...^{4} = 238 \\ b = {238 \over 0{,}957...^{4}} ≈ 284\end{matrix}\)

\(y = 284 ⋅ 0{,}957^{x} \text{.}\)

\(y = b ⋅ g^{x}\) met \(g = ({384 \over 328})^{{1 \over 9 - 4}} = 1{,}032...\)

\(\begin{rcases}y = b ⋅ 1{,}032...^{x} \\ x = 4 \text{ en } y = 328\end{rcases} \begin{matrix}b ⋅ 1{,}032...^{4} = 328 \\ b = {328 \over 1{,}032...^{4}} ≈ 289\end{matrix}\)

\(y = 289 ⋅ 1{,}032^{x} \text{.}\)

Rasterpunten \((2 , 600)\) en \((8 , 8\,000)\) aflezen.

\(y = b ⋅ g^{x}\) met \(g = ({8\,000 \over 600})^{{1 \over 8 - 2}} = 1{,}539...\)

\(\begin{rcases}y = b ⋅ 1{,}539...^{x} \\ x = 2 \text{ en } y = 600{,}00\end{rcases} \begin{matrix}b ⋅ 1{,}539...^{2} = 600{,}00 \\ b = {600{,}00 \over 1{,}539...^{2}} \\ b = 253{,}029...\end{matrix}\)

\(y = 253 ⋅ 1{,}540^{x} \text{.}\)

\({32{,}62 \over 41{,}82} ≈ 0{,}78\)

\({25{,}44 \over 32{,}62} ≈ 0{,}78\)
\({19{,}85 \over 25{,}44} ≈ 0{,}78\)
\({15{,}48 \over 19{,}85} ≈ 0{,}78\)

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

\(y = b ⋅ g^{x}\) met \(g = 0{,}78\)

\(b\) is de waarde bij \(x = 0 \text{,}\) dus \(b = 41{,}82 \text{.}\)

Dus \(y = 41{,}82 ⋅ 0{,}78^{x} \text{.}\)

\(g = ({58{,}87 \over 26{,}82})^{{1 \over 2\,018 - 2\,012}} ≈ 1{,}14\)

\(g = ({76{,}51 \over 58{,}87})^{{1 \over 2\,020 - 2\,018}} ≈ 1{,}14\)
\(g = ({147{,}31 \over 76{,}51})^{{1 \over 2\,025 - 2\,020}} ≈ 1{,}14\)

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

\(y = b ⋅ g^{x}\) met \(g = 1{,}14\)

\(\begin{rcases}y = b ⋅ 1{,}14^{x} \\ x = 4 \text{ en } y = 26{,}82\end{rcases} \begin{matrix}b ⋅ 1{,}14^{4} = 26{,}82 \\ b = {26{,}82 \over 1{,}14^{4}} \\ b ≈ 15{,}88\end{matrix}\)

Dus \(y = 15{,}88 ⋅ 1{,}14^{x} \text{.}\)

\(y = a x + b\) met \(a = {\Delta y \over \Delta x} = {5 - 11 \over 2 - 1} = -6\)

\(\begin{rcases}y = -6 x + b \\ \text{door } (1 , 11)\end{rcases} \begin{matrix}-6 ⋅ 1 + b = 11 \\ -6 + b = 11 \\ b = 17\end{matrix}\)

Dus \(l{:}\,y = -6 x + 17\)

\(y = b ⋅ g^{x}\) met \(g = ({5 \over 11})^{{1 \over 2 - 1}} = 0{,}454...\)

\(\begin{rcases}y = b ⋅ 0{,}454...^{x} \\ \text{door } (1 , 11)\end{rcases} \begin{matrix}11 = b ⋅ 0{,}454...^{1} \\ b = {11 \over 0{,}454...^{1}} \\ b = 24{,}2\end{matrix}\)

Dus \(y = 24{,}2 ⋅ 0{,}455^{x} \text{.}\)

\({19{,}10 \over 25{,}46} ≈ 0{,}75\)

\({14{,}32 \over 19{,}10} ≈ 0{,}75\)
\({10{,}74 \over 14{,}32} ≈ 0{,}75\)

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 \(b = 25{,}46 \text{.}\)

Dus \(y = 25{,}46 ⋅ 0{,}75^{x} \text{.}\)

\({\Delta y \over \Delta x} = {19{,}02 - 17{,}40 \over 2\,016 - 2\,010} = 0{,}27\)

\({\Delta y \over \Delta x} = {20{,}37 - 19{,}02 \over 2\,021 - 2\,016} = 0{,}27\)
\({\Delta y \over \Delta x} = {20{,}91 - 20{,}37 \over 2\,023 - 2\,021} = 0{,}27\)
\({\Delta y \over \Delta x} = {21{,}18 - 20{,}91 \over 2\,024 - 2\,023} = 0{,}27\)

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

\(y = a x + b\) met \(a = 0{,}27\)

\(\begin{rcases}y = 0{,}27 x + b \\ x = 4 \text{ en } y = 17{,}4\end{rcases} \begin{matrix}0{,}27 ⋅ 4 + b = 17{,}4 \\ 1{,}08 + b = 17{,}4 \\ b = 16{,}32\end{matrix}\)

Dus \(y = 0{,}27 x + 16{,}32\)

Punt \(\text{A} (1 , 800\,000) \text{.}\)

Punt \(\text{B} (4 , 900) \text{.}\)

Punt \(\text{C} (7 , 40\,000) \text{.}\)

Rasterpunten \((10 , 2)\) en \((80 , 5)\) aflezen.

\(\log(y) = a x + b\) met \(a = {\Delta \log(y) \over \Delta x} = {5 - 2 \over 80 - 10} = \frac{3}{70}\)

\(\begin{rcases}\log(y) = \frac{3}{70} x + b \\ \text{door } (10 , 2)\end{rcases} \begin{matrix}\frac{3}{70} ⋅ 10 + b = 2 \\ \frac{30}{70} + b = 2 \\ b = 1\frac{4}{7}\end{matrix}\)

\(\log(y) = \frac{3}{70} x + 1\frac{4}{7}\)
\(y = 10^{\frac{3}{70} x + 1\frac{4}{7}}\)

\(y = 10^{\frac{3}{70} x} ⋅ 10^{1\frac{4}{7}}\)
\(\text{ } = 10^{1\frac{4}{7}} ⋅ (10^{\frac{3}{70}})^{x}\)
\(\text{ } = 37 ⋅ 1{,}104^{x}\)

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

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

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

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

\(y = 10^{3\frac{2}{3}} ⋅ x^{-\frac{1}{3}}\)
\(y = 4\,642 ⋅ x^{-0{,}33}\)

De grafiek hoort bij een lineair verband.

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