A population of bacteria is growing according to the equation
P(t) = 1650e^0.22t. Estimate when the population will about 5656.
I think oobleck dropped a decimal place , it should be ....
1650 e^0.22t = 5656
e^0.22t = 3.4278
0.22t = ln 3.4278 = 1.23194...
t = 5.5997
If t is in years, it would take appr 5.6 years
nice save -- thx
To estimate when the population will reach about 5656, we need to find the value of t for which the population function P(t) equals 5656.
The given population function is P(t) = 1650e^(0.22t).
We can set up the equation and solve for t as follows:
5656 = 1650e^(0.22t)
To isolate the exponential term, divide both sides of the equation by 1650:
5656/1650 = e^(0.22t)
Now, take the natural logarithm (ln) of both sides to remove the exponential:
ln(5656/1650) = 0.22t
Next, divide both sides of the equation by 0.22 to solve for t:
t = ln(5656/1650) / 0.22
Using a calculator, evaluate the right side of the equation to find the estimated value of t.
165 e^0.22t = 5656
e^0.22t = 34.278
0.22t = ln 34.278 = 3.535
t = 16.066