If there are 1,000 bugs at the start of an experiment and 1800 after 1 day, how long until there are 10,000?

Answer 1

The general equation for exponential growth is

A=Ao(e^kt)

A = Final amount

Ao = Initial amount

k = growth constant

t = time

Plug in the variables known

1800=1000(e^k(1)) We can solve this for k to find the growth constant. Divide both sides of the equation by 1000

1.8=e^k Take the ln of both sides

ln(1.8)=ln(e^k) The exponent comes to the front

ln(1.8)=(k)(ln(e) ln e = 1

ln(1.8)=k

k=0.5878

Now, can use k to solve for t in the second part of this problem. Plug in the values you know. You will solve for t.

10000=1000(e^0.5878t) Divide both sides of the equation by 1000

10=(e^0.5878t) Take the ln of both sides

ln(10)=ln(e^0.5878t) The exponent comes to the front

ln(10)=(0.5878t) ln(e) ln e = 1

ln(10)=0.5878t Divide both sides by 0.5878 and simplify

t=3.9 days

Answer 2

800 per day

y = mx + b and m=800 with (0,1000)

so y = 800x + 1000

10000 = 800x + 1000

9000 = 800x

9000/800 = x

90/8 = x

x = 45/4 = 11 and 1/4

11 days and 6 hours