What is the Exponential Growth Rate of Cells under a Microscope?

[iScience]

Homework Statement

There are initially two cells seen through a microscope. every hour, a computer was tasked with taking a cell count through the microscope. Taking the sum of all the cell counts after the 7th hour, the total cell count was 280 cells. Find the exponential growth rate.

Homework Equations

$$P = P_0\sum_{t = 1}^{7}{e^{kt}}$$
$$P_0 = 2$$

The Attempt at a Solution

$$2\sum_{t=1}^{7}e^{kt} = P$$
$$\sum_{t=1}^{7}e^{kt} = 280/2$$
$$ln(140) = \sum_{t=1}^{7}(kt) = k(\sum_{t=1}^{7}t) = k*(28)$$
$$k = \frac{ln(140)}{28}$$
$$ k = 0.176487$$

if you plug this into the original equation:

$$P_0\sum_{t = i}^{7}{e^{0.176487t}}$$
this sum from 1 to 7 does not equal my final P value, what am i missing??

 

[haruspex]

The log of a sum is not the sum of the logs. You need to sum the series first.

 

[SammyS]

Homework Statement

There are initially two cells seen through a microscope. every hour, a computer was tasked with taking a cell count through the microscope. Taking the sum of all the cell counts after the 7th hour, the total cell count was 280 cells. Find the exponential growth rate.

Homework Equations

$$P = P_0\sum_{t = 1}^{7}{e^{kt}}$$ $$P_0 = 2$$

The Attempt at a Solution

[/B]$$2\sum_{t=1}^{7}e^{kt} = P$$ $$\sum_{t=1}^{7}e^{kt} = 280/2$$ $$ln(140) = \sum_{t=1}^{7}(kt) = k(\sum_{t=1}^{7}t) = k*(28)$$ $$k = \frac{ln(140)}{28}$$ $$ k = 0.176487$$
if you plug this into the original equation:
$$P_0\sum_{t = i}^{7}{e^{0.176487t}}$$
this sum from 1 to 7 does not equal my final P value, what am i missing??

The first "Relevant Equation" $\displaystyle \ P = P_0\sum_{t = 1}^{7}{e^{kt}} \$ is incorrect for exponential growth.

It's $\displaystyle \ P(t) = P_0\ e^{kt} \$

 

[haruspex]

The first "Relevant Equation" $\displaystyle \ P = P_0\sum_{t = 1}^{7}{e^{kt}} \$ is incorrect for exponential growth.

It's $\displaystyle \ P(t) = P_0\ e^{kt} \$

I believe iScience is using P for the cumulative total of observations, not the population at a time step.

 

[Ray Vickson]

Homework Statement

There are initially two cells seen through a microscope. every hour, a computer was tasked with taking a cell count through the microscope. Taking the sum of all the cell counts after the 7th hour, the total cell count was 280 cells. Find the exponential growth rate.

Homework Equations

$$P = P_0\sum_{t = 1}^{7}{e^{kt}}$$
$$P_0 = 2$$

The Attempt at a Solution

$$2\sum_{t=1}^{7}e^{kt} = P$$
$$\sum_{t=1}^{7}e^{kt} = 280/2$$
$$ln(140) = \sum_{t=1}^{7}(kt) = k(\sum_{t=1}^{7}t) = k*(28)$$
$$k = \frac{ln(140)}{28}$$
$$ k = 0.176487$$

if you plug this into the original equation:

$$P_0\sum_{t = i}^{7}{e^{0.176487t}}$$
this sum from 1 to 7 does not equal my final P value, what am i missing??

As has already been pointed out: you cannot distribute the "log" over the "sum".
Your basic problem is to solve the equation
2(r + r^2 + \cdots + r^7) = 280, \; (r = e^k)
You can solve for $r$ first, then get $k$ from that. However, solving for $r$ involves solving a 7th degree polynomial (and even doing the 7-term sum first does not help), so you must fall back on graphical/numerical solution methods. There will not be any nice formulas you can apply.

 

[Chestermiller]

As has already been pointed out: you cannot distribute the "log" over the "sum".
Your basic problem is to solve the equation
2(r + r^2 + \cdots + r^7) = 280, \; (r = e^k)
You can solve for $r$ first, then get $k$ from that. However, solving for $r$ involves solving a 7th degree polynomial (and even doing the 7-term sum first does not help), so you must fall back on graphical/numerical solution methods. There will not be any nice formulas you can apply.

This is the sum of a geometric progression. That might simplify things a little.

 

[Ray Vickson]

This is the sum of a geometric progression. That might simplify things a little.

As I said in # 5, even doing the sum does not help much; it replaces a 7th degree polynomlal equation by an 8th degree one!

 

[Chestermiller]

As I said in # 5, even doing the sum does not help much; it replaces a 7th degree polynomlal equation by an 8th degree one!

But, the 8th degree equation only involves r⁸ and r. I solved the equation $$\frac{r^8-1}{r-1}=140$$ in 4 iterations. My first guess was r = 1.75. So, my sequence of guesses was
1.75
1.8
1.81
1.805

This all took a total of less than 5 minutes on my calculator.

Chet

 

[Ray Vickson]

But, the 8th degree equation only involves r⁸ and r. I solved the equation $$\frac{r^8-1}{r-1}=140$$ in 4 iterations. My first guess was r = 1.75. So, my sequence of guesses was
1.75
1.8
1.81
1.805

This all took a total of less than 5 minutes on my calculator.

Chet

Well, of course there are numerous ways of solving that problem (or the non-summed one) numerically, and some are faster and/or easier than others. The main point I wanted to get across to the OP was that some such method must be used, and that there will be no nice "closed-form" formula that can be used to get the solution. However, it looks like he/she may have abandoned the thread.

 

[Chestermiller]

Well, of course there are numerous ways of solving that problem (or the non-summed one) numerically, and some are faster and/or easier than others. The main point I wanted to get across to the OP was that some such method must be used, and that there will be no nice "closed-form" formula that can be used to get the solution. However, it looks like he/she may have abandoned the thread.

Thanks Ray. Anyway, it's a good thing he left because I summed the geometric progression incorrectly anyway. I should have solved:
$$\frac{r(r^7-1)}{r-1}=140$$

Oh well.