Question

Help please

A population of bacteria is growing according to the equation p(t)=800e^0.14t Estimate when the population will exceed 1151.

t= ---------

Answer 1

t=2.59839

Explanation:

You are trying to solve for t in the equation

1151=800*e^(.14t) --> 1.43875=e^(.14t)

to get rid of the e, you take the natural log of both sides, denoted by "ln" on calculators. The natural log of e to the power of something always equals the something

ln(1.43875)=ln(e^(.14t)) --> .36377=.14t --> 2.59839=t

Answer 2

To find the value of t when the population exceeds 1151, we need to solve the equation:

1151 = 800e^(0.14t)

First, divide both sides of the equation by 800:

1.43875 = e^(0.14t)

Now, take the natural logarithm (ln) of both sides:

ln(1.43875) = ln(e^(0.14t))

Using the property of logarithms, we can move the exponent in front of the logarithm:

ln(1.43875) = 0.14t * ln(e)

Since ln(e) = 1:

ln(1.43875) = 0.14t

Now, divide both sides by 0.14:

t = ln(1.43875) / 0.14

Finally, calculate the value of t:

t ≈ 2.71

So, the population will exceed 1151 at approximately t = 2.71.

Explanation: