# Math Help - exp growth

1. ## exp growth

The count in a bacteria culture was 900 after 10 minutes and 1400 after 35 minutes.
What was the initial size of the culture?
doubling period = ?
population after 110 minutes = ?
When will the population reach 15000?

can someone explain to me on how to do this problem?

2. Hello, viet!

The count in a bacteria culture was 900 after 10 minutes and 1400 after 35 minutes.
(a) What was the initial size of the culture?
(b) What is the doubling period?
(c) What is the population after 110 minutes?
(d) When will the population reach 15000?

We assume an exponential growth function: . $P(t) \:=\:ae^{bt}$
. . and determine $a$ and $b$ from the given data.

. . $\begin{array}{cc}P(10) = 900 \\ P(35) = 1400\end{array}
\begin{array}{cc} \Rightarrow \\ \Rightarrow\end{array}
\begin{array}{cc}ae^{10b} \,=\,900 \\ ae^{35b} \,=\,1400 \end{array}
\begin{array}{cc}(1)\\(2)\end{array}$

Divide (2) by (1): . $\frac{ae^{35b}}{ae^{10b}} \:=\:\frac{1400}{900}\quad\Rightarrow\quad e^{25b}\,=\,\frac{14}{9}\quad\Rightarrow\quad 25b \,=\,\ln\left(\frac{14}{9}\right)$

. . Hence: . $b \:=\:\frac{1}{25}\ln\left(\frac{14}{9}\right) \:\approx\:0.0177$

Substitute into (1): . $ae^{0.0177(10)} \:=\:900\quad\Rightarrow\quad a \:=\:\frac{900}{e^{0.177}} \:\approx\:754$

The function is: . $\boxed{P(t)\:=\:754e^{0.0177t}}$

(a) When $t = 0\!:\;\;P(0)\:=\:754e^0\quad\Rightarrow\quad\boxed { P(0) \,=\,754}$

(b) If $P(t) = 1508\!:\;\;754e^{0.0177t} \:=\:1508\quad\Rightarrow\quad e^{0.0117t}\,=\,2\quad\Rightarrow\quad 0.0117t\,=\,\ln 2$

. . $t \:=\:\frac{\ln2}{0.0177} \:\approx\:\boxed{39.16\text{ minutes}}$

(c) $P(110)\:=\:574e^{0.0177(110)} \:\approx\:\boxed{4022.38}$

(d) for $P(t) = 15,000\!:\;\;574e^{0.0177t} \,=\,15,000\quad\Rightarrow\quad e^{0.0177t} \,=\,\frac{15,000}{574}$

. . $0.0177t \,=\,\ln\left(\frac{15,000}{574}\right)\quad\Right arrow\quad t \:=\:\frac{1}{0.0177}\ln\left(\frac{15,000}{574}\r ight)\:\approx\:\boxed{184.36\text{ minutes}}$

3. Originally Posted by viet
The count in a bacteria culture was 900 after 10 minutes and 1400 after 35 minutes.
What was the initial size of the culture?
doubling period = ?
population after 110 minutes = ?
When will the population reach 15000?

can someone explain to me on how to do this problem?
Here is one way.

his problem?

Let x = initial size of the bacteria culture
And r = common ratio.

You know geometric progression?
a1 = x
a2 = x*r
a3 = (x*r)*r = x*(r^2)
a4 = x(r^2)*r = x(r^3)
.
.
an = x(r^(n-1)) -------------***

After 10 minutes,
900 = x*r^(10-1)
900 = x(r^9) ----------------(i)

After 35 min,
1400 = x(r^34) ---------------(ii)

(ii) divided by (i),
1400/900 = r^(34 -9)
14/9 = r^25
r = (14/9)^(1/25) = 1.01783 -----**

Substitute that into, say, (i),
900 = x(1.01783)^9
x = 900 /(1.01783)^9 = 676.6536
Or, initial size of culture = 677 numbers ----------answer.

-----------------------------------------------------------------
The doubling period...

I understand that as when will the culture double in size.
So,
2(677) = 677*(1.01783)^(y-1)
2 = (1.01783)^(y-1)
Take common logs of both sides,
log(2) = (y-1)log(1.01783)
y-1 = log(1.01783) /log(2)
y -1 = 0.0255
y = 1.0255 ----------------the initial size doubles after 1.0255 min.

If you analyze the computations above, or since the size of the culture is cancelled in the

second line, thus the size doesn't matter, then it means every 1.0255 minutes the size

doubles.
Therefore, doubling period = ervery 1.0255 min. ----------answer.

---------------------------------------------------------------------
population after 110 minutes = ?

an = a1*r^(n-1)

a(110) = 677*(1.01783)^109 = 4647 numbers ------------answer.

----------------------------
When will the population reach 15000?

15000 = 677*(1.01783)^(z-1)
15000/677 = 1.01783^(z-1)
log(15000/677) = (z-1)log(1.01783)
z-1 = log(15000/677) / log(1.01783)
z -1 = 175.304, or 175
z = 176

Therefore, the culture will be 15000 numbers in 176 min. -----------answer.