Math

QuestionA bacterial colony of 1000 doubles in 12 hours. Find growth rate kk and time to reach 200,000 bacteria.

Studdy Solution

STEP 1

Assumptions1. The initial number of bacteria (Q0) is1000. . The number of bacteria doubles every12 hours.
3. The growth of the bacteria follows the exponential function Q=Q0ektQ=Q_{0} e^{k t}, where QQ is the final number of bacteria, Q0Q_{0} is the initial number of bacteria, kk is the growth rate, and tt is the time.

STEP 2

First, we need to find the growth rate kk. We know that the number of bacteria doubles every12 hours, so we can set Q=2Q0Q =2Q_{0} and t=12t =12 in the exponential function and solve for kk.
2Q0=Q0ek122Q_{0} = Q_{0} e^{k \cdot12}

STEP 3

Divide both sides of the equation by Q0Q_{0} to isolate the exponential term on the right side.
2=ek122 = e^{k \cdot12}

STEP 4

Take the natural logarithm of both sides to solve for kk.
ln(2)=k12\ln(2) = k \cdot12

STEP 5

Divide both sides by12 to find the value of kk.
k=ln(2)12k = \frac{\ln(2)}{12}

STEP 6

Calculate the value of kk to four decimal places.
k0.0578k \approx0.0578

STEP 7

Now that we have the growth rate, we can find out how long it takes for the bacteria to grow to200,000. We set Q=200000Q =200000 in the exponential function and solve for tt.
200000=1000e0.057t200000 =1000 e^{0.057 t}

STEP 8

Divide both sides by1000 to isolate the exponential term on the right side.
200=e0.0578t200 = e^{0.0578 t}

STEP 9

Take the natural logarithm of both sides to solve for tt.
ln(200)=.0578t\ln(200) =.0578 t

STEP 10

Divide both sides by0.0578 to find the value of tt.
t=ln(200)0.0578t = \frac{\ln(200)}{0.0578}

STEP 11

Calculate the value of tt.
t62.3hourst \approx62.3 \, hoursIt takes approximately62.3 hours for the bacteria to grow to200,000.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord