Question

A biologist is researching a newly-discovered species of bacteria. At time t = 0 hours, he puts two hundred bacteria into what he has determined to be a favorable growth medium. Eight hours later, he measures 1200 bacteria. Assuming exponential growth, what is the growth constant "k" for the bacteria? (Round k to three decimal places.)

Answer 1

You can use the exponential equation
$$
\Large y = A_o e^{kt}
$$

Answer 2

how do i use it?

Answer 3

\( \large A_0 \) is the population at time zero

Answer 4

which would be 200?

Answer 5

@perl

Answer 6

@mathteacher1729 @mathmath333 @amistre64

Answer 7

yes that is correct

Answer 8

how would I find e and kt?

Answer 9

e is a special constant, it stands for 2.718182...

Answer 10

k is what we are looking for right

Answer 11

would t be 0

Answer 12

It says 8 hours later, then population is 1200

$$\Large { y = 200e^{kt}
\\ \therefore \\
1200 = 200 e^{k\cdot 8}
\\ \text{solve for k }

}
$$

Answer 13

no put \(t=8\)

Answer 14

do i divide both sides by 200?

Answer 15

then id be left with 6= e^k(8)

Answer 16

@perl

Answer 17

@mathmath333 could you help

Answer 18

u have to take natural log \(\large \ln\) on both sides now

Answer 19

I got 12000 = 2,000 ^k(8)

Answer 20

\(\large \color{black}{\begin{align} e^{8k}&=6\hspace{.33em}\\~\\
\ln e^{8k}&=\ln 6\hspace{.33em}\\~\\
8k\ln e&=\ln 6\hspace{.33em}\\~\\
8k&=\ln 6\hspace{.33em}\\~\\
k&=\dfrac{\ln 6}{8}\hspace{.33em}\\~\\
\end{align}}\)

Answer 21

is that the final answer?

Answer 22

use

\(\large \color{black}{\begin{align}
\ln 6 \approx 1.79\hspace{.33em}\\~\\
\end{align}}\)

Answer 23

\(\large \color{black}{\begin{align}
k=\dfrac{1.79}{8}=?\hspace{.33em}\\~\\
\end{align}}\)

Answer 24

0.22375

Answer 25

yep correct

Answer 26

upto 3 decimal is enough

Answer 27

was the first part of the process correct?

Answer 28

Great thanks! could I bother you with one more question?

Answer 29

\(\large \color{black}{\begin{align}
6=e^{8k}\hspace{.33em}\\~\\
\end{align}}\) is coorect

Answer 30

yea bring it on

Answer 31

Brad created a chart that shows the population of a town will increase to 109,627 people from a current population of 15,200 people. The rate of increase is an annual increase of 6.52%. Brad forgot to include the number of years this increase will take. How many years was it?
(Solve algebraically.)

Answer 32

use the same formula

Answer 33

\(\large \color{black}{\begin{align}
y = A_o e^{kt}\hspace{.33em}\\~\\
y=109,627\hspace{.33em}\\~\\
A_o=15,200\hspace{.33em}\\~\\
k=6.52\hspace{.33em}\\~\\
\end{align}}\)

Answer 34

so would i divide both sides by 15,200?

Answer 35

@mathmath333

Answer 36

\(\large \color{black}{\begin{align} y &= A_o e^{kt}\hspace{.33em}\\~\\
109627&= 15200 e^{6.52t}\hspace{.33em}\\~\\
\dfrac{109627}{15200}&= e^{6.52t}\hspace{.33em}\\~\\
7.21&= e^{6.52t}\hspace{.33em}\\~\\
\ln 7.21&= \ln e^{6.52t}\hspace{.33em}\\~\\
\ln 7.21&= 6.52t\ln e\hspace{.33em}\\~\\
\ln 7.21&= 6.52t\hspace{.33em}\\~\\
t&=\dfrac{\ln 7.21}{6.52}\hspace{.33em}\\~\\
t&=0.302\hspace{.33em}\\~\\
\end{align}}\)

Answer 37

u can use wolfram for complex calculations
https://www.wolframalpha.com/