Question

use regression to find an exponential equation that best fits the data above. the equation has form y = ab^x where: a = b = round your answers to 2 decimal places. question help: post to forum submit question jump to answer

Explanation:

Step1: Take natural - log of both sides of $y = ab^{x}$

If $y = ab^{x}$, then $\ln(y)=\ln(a)+x\ln(b)$. Let $Y = \ln(y)$, $A=\ln(a)$ and $B = \ln(b)$. The equation becomes $Y=A + Bx$, which is a linear equation.

Step2: Calculate $Y$ values

For $x = 1,y = 782$, $Y_1=\ln(782)\approx6.66$.
For $x = 2,y = 1272$, $Y_2=\ln(1272)\approx7.15$.
For $x = 3,y = 2154$, $Y_3=\ln(2154)\approx7.68$.
For $x = 4,y = 3536$, $Y_4=\ln(3536)\approx8.17$.
For $x = 5,y = 5472$, $Y_5=\ln(5472)\approx8.61$.
For $x = 6,y = 9167$, $Y_6=\ln(9167)\approx9.12$.

Step3: Use linear - regression formula for $Y = A + Bx$

The formulas for the least - squares regression line $Y = A + Bx$ are:
$B=\frac{n\sum_{i = 1}^{n}x_{i}Y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}Y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}$ and $A=\overline{Y}-B\overline{x}$
Here, $n = 6$, $\sum_{i=1}^{6}x_{i}=1 + 2+3 + 4+5 + 6=21$, $\sum_{i = 1}^{6}Y_{i}=6.66 + 7.15+7.68+8.17+8.61+9.12 = 47.39$, $\sum_{i = 1}^{6}x_{i}^{2}=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}=91$, $\sum_{i = 1}^{6}x_{i}Y_{i}=1\times6.66+2\times7.15+3\times7.68+4\times8.17+5\times8.61+6\times9.12=189.95$
$B=\frac{6\times189.95-21\times47.39}{6\times91 - 21^{2}}=\frac{1139.7-995.19}{546 - 441}=\frac{144.51}{105}\approx1.38$
$\overline{x}=\frac{21}{6}=3.5$, $\overline{Y}=\frac{47.39}{6}\approx7.90$
$A=7.90-1.38\times3.5=7.90 - 4.83=3.07$

Step4: Find $a$ and $b$

Since $A=\ln(a)$, then $a = e^{A}$. Since $A\approx3.07$, $a = e^{3.07}\approx21.50$
Since $B=\ln(b)$, then $b = e^{B}$. Since $B\approx1.38$, $b = e^{1.38}\approx3.98$

Answer:

$a\approx21.50$
$b\approx3.98$