1. ## logs

Please can you explain how do I do the following:

1. logN1 = log(N0 multiplied by 2^n)

Make n the subject?

All logarithms must be kept to the base 10.

Formula: Growth rate (k)

k = logN1 - logN0 / tlog2

2. A population of Escherichia coli, which increases from 2000 cells to 120 000 cells in 124 minutes during exponential phase. When incubated at 40 degrees.

2.1 Calc. the number of generations per hour? convert minutes to hours.

I am not particular sure how to use these formula, and calc. logs on the calculator.

3.1 Make t the subject of the formula:

g = tlog2 / logN1 - logN0

2. Originally Posted by ticktack
Please can you explain how do I do the following:

1. logN1 = log(N0 multiplied by 2^n)

Make n the subject?
recall that $\log_a (xy) = \log_a x + \log_a y$ and $\log_a \left( \frac xy \right) = \log_a x - \log_a y$

So, we have $\log N_1 = \log (N_0 \cdot 2^n)$

$\Rightarrow \log N_1 = \log N_0 + \log 2^n$

$\Rightarrow \log 2^n = \log N_1 - \log N_0$

$\Rightarrow n \log 2 = \log \left( \frac {N_1}{N_0} \right)$ .....since $\log_a \left(x^n \right) = n \log_a x$

$\Rightarrow n = \frac {\log \left( \frac {N_1}{N_0} \right)}{\log 2}$

3. Originally Posted by ticktack

Formula: Growth rate (k)

k = logN1 - logN0 / tlog2
did you mean $\frac {\log N_1 - \log N_0}{t \log 2}$??

4. yes thats what I meant Jhevon..

5. Could you please tell me if the following working out is correct..

Formula: Growth rate (k)

k = logN1 - logN0 / tlog2

2. A population of Escherichia coli, which increases from 2000 cells to 120 000 cells in 124 minutes during exponential phase. When incubated at 40 degrees.

2.1 Calc. the number of generations per hour? convert minutes to hours.

k = logN1 - logN0 / tlog2

N1=120000 N0=2000 t=2hrs 4min is 2.4

k = log12.10^4 - log2.10^2 / 2.4log2

=4-2 / 2.4log2

=2,8 generations per hour

3.1 Make t the subject of the formula:
g = tlog2 / logN1 - logN0

glogN1-logN0 / log2=t