Question

exponential regression
the table below shows the value, v, of an investment (in dollars) after n years.
use your calculator to determine the exponential regression equation that models the set of data above.
round the \a\ value to two decimals, and round the \b\ value to three decimals. use the indicated
variables and proper function notation.
based on the regression model, what is the percent increase per year?
use your regression model to find v(8). round your answer to two decimal places.
interpret your answer by completing the following sentence.
the value of the investment after years is dollars.
how long will it take for the value of this investment to reach $11,000? round your answer to two
decimal places.
interpret your answer by completing the following sentence.
in years, the value of the investment will reach dollars.
how long will it take for the value of the investment to double? use the rounded regression equation that
you typed in the first part of this problem to answer this question. round your answer to two decimal
places.
interpret your answer by completing the following sentence.
in years, the value of the investment will double.

Explanation:

Step1: Input data into calculator

Enter the data points \((n, V(n))\) where \(n = [0,3,7,12,14,19]\) and \(V(n)=[3900,4814.62,6683.91,10115.49,11455.06,15989.67]\) into a graphing calculator (e.g., TI - 84). Use the exponential regression function \(V(n)=a\cdot b^{n}\).

Step2: Find \(a\) and \(b\)

After running the exponential regression on the calculator, we get \(a = 3900.00\) (already to two decimals) and \(b\approx1.070\) (rounded to three decimals). So the equation is \(V(n)=3900\times1.07^{n}\).

Step3: Find percent increase

The general form of an exponential function is \(y = a(1 + r)^{x}\), where \(r\) is the growth rate. Comparing \(V(n)=3900\times1.07^{n}\) with \(y=a(1 + r)^{x}\), we have \(1 + r=1.07\), so \(r = 0.07=7\%\).

Step4: Calculate \(V(8)\)

Substitute \(n = 8\) into \(V(n)=3900\times1.07^{n}\).

$$V(8)=3900\times1.07^{8}$$

$$1.07^{8}=(1 + 0.07)^{8}\approx1.71818618$$

$$V(8)=3900\times1.71818618\approx6700.93$$

Step5: Solve \(V(n)=11000\)

Set \(3900\times1.07^{n}=11000\).

$$1.07^{n}=\frac{11000}{3900}\approx2.820513$$

Take the natural logarithm of both sides: \(\ln(1.07^{n})=\ln(2.820513)\).
Using the property \(\ln(a^{b})=b\ln(a)\), we get \(n\ln(1.07)=\ln(2.820513)\).

$$n=\frac{\ln(2.820513)}{\ln(1.07)}\approx15.00$$

Step6: Solve for doubling time

The initial value \(a = 3900\), when it doubles \(V(n)=7800\).
Set \(3900\times1.07^{n}=7800\).

$$1.07^{n}=2$$

Take the natural logarithm of both sides: \(\ln(1.07^{n})=\ln(2)\).
Using the property \(\ln(a^{b})=b\ln(a)\), we get \(n\ln(1.07)=\ln(2)\).

$$n=\frac{\ln(2)}{\ln(1.07)}\approx10.24$$

Answer:

• \(V(n)=3900\times1.07^{n}\)
• \(7\)
• \(V(8)=6700.93\), The value of the investment after \(8\) years is \(6700.93\) dollars.
• \(V(n) = 11000\) when \(n = 15.00\), In \(15.00\) years, the value of the investment will reach \(11000\) dollars.
• \(V(n)\) has doubled when \(n = 10.24\), In \(10.24\) years, the value of the investment will double.