1. ## exponential functions

1.Tiger drops a golf ball from a height of 60 m. After each bounce, the height of the balls bounce is 4/5 the height of the previous bounce. How high does the ball bounce after the tenth bounce? (Set up an exponential equation for the height.)

2.The deer population on the Bruce trail in 2005, 2006, and 2007 was recorded as 225, 270, and 324 respectively. Using exponential functions, predict the number of deer in 2015

2. Originally Posted by william
1.Tiger drops a golf ball from a height of 60 m. After each bounce, the height of the balls bounce is 4/5 the height of the previous bounce. How high does the ball bounce after the tenth bounce? (Set up an exponential equation for the height.)
What distance is covered in the initial drop?

What is 4/5 of this distance? (This is the height reached after the first bounce.)

What is 4/5 of this first-bounce distance? (This is the height reached after the second bounce.)

Keep going until you see a pattern or until you reach the value for after the tenth bounce.

Originally Posted by william
2.The deer population on the Bruce trail in 2005, 2006, and 2007 was recorded as 225, 270, and 324 respectively. Using exponential functions, predict the number of deer in 2015
Are you supposed to use statistical software to find an exponential regression for this?

3. Originally Posted by stapel
What distance is covered in the initial drop?

What is 4/5 of this distance? (This is the height reached after the first bounce.)

What is 4/5 of this first-bounce distance? (This is the height reached after the second bounce.)

Keep going until you see a pattern or until you reach the value for after the tenth bounce.

Are you supposed to use statistical software to find an exponential regression for this?

Thanks

1. how do i set up the exponential equation for that?

2. no graphing software should be needed

4. Originally Posted by william
1. how do i set up the exponential equation for that?
That's the pattern you're looking for.

Originally Posted by william
2. no graphing software should be needed
Okay. Start by setting up variables so the first data point represents the initial value at "t = 0". Then take whatever equation form they've given you (different books use different forms), and plug the other two points into this. You will then have two equations (with the input t-value and the output value already plugged in) in two unknowns, being the growth constant and the base of the exponential.

Solve the system for the unknowns. Then rewrite the equation form with the initial value, the growth constant, and the base plugged in.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!

5. Hello, william;!

1.Tiger drops a golf ball from a height of 60 m.
After each bounce, the height of the balls bounce is 4/5 the height of the previous bounce.
How high does the ball bounce after the tenth bounce?
(Set up an exponential equation for the height.)

The height $h$ at the $n^{th}$ bounce is given by: . $h(n) \:=\:60\left(\frac{4}{5}\right)^n$

2.The deer population on the Bruce trail in 2005, 2006, and 2007 was recorded as
225, 270, and 324 respectively.
Using exponential functions, predict the number of deer in 2015
We have: . $\begin{array}{|c|c|} \hline \text{Year} & \text{Pop'n} \\ \hline \hline 2005 & 225 \\ \hline 2006 & 270 \\ \hline
2007 & 324 \\ \hline \end{array}$

Take consecutive ratios: . $\begin{array}{ccc}\frac{270}{225} &=& \frac{6}{5} \\ \\[-4mm] \frac{324}{270} &=& \frac{6}{5} \end{array}$

Each population is $\tfrac{6}{5}$ times the previous year's popullation.

The population $n$ years after 2005 is given by: . $P(n) \:=\:225\left(\frac{6}{5}\right)^{n-1}$

In 2015 $(n = 10)\!:\;\;P(10) \:=\:225\left(\frac{6}{5}\right)^9 \:\approx\;1161$ deer.

6. Originally Posted by Soroban
Hello, william;!

The height $h$ at the $n^{th}$ bounce is given by: . $h(n) \:=\:60\left(\frac{4}{5}\right)^n$

We have: . $\begin{array}{|c|c|} \hline \text{Year} & \text{Pop'n} \\ \hline \hline 2005 & 225 \\ \hline 2006 & 270 \\ \hline
2007 & 324 \\ \hline \end{array}$

Take consecutive ratios: . $\begin{array}{ccc}\frac{270}{225} &=& \frac{6}{5} \\ \\[-4mm] \frac{324}{270} &=& \frac{6}{5} \end{array}$

Each population is $\tfrac{6}{5}$ times the previous year's popullation.

The population $n$ years after 2005 is given by: . $P(n) \:=\:225\left(\frac{6}{5}\right)^{n-1}$

In 2015 $(n = 10)\!:\;\;P(10) \:=\:225\left(\frac{6}{5}\right)^9 \:\approx\;1161$ deer.

Soroban you are a great mathematician, every response from you is top notch.

Thank you to both.