Question

Calling Genius Mathematicians!!!!!! Please Help!!!!!!! 7 and 8!!!!!!

Answer 1

7) We are given
`G=Ae^(0.6t)`. Recall that e is a real, albeit irrational, number, and it is given that G is the final number of bacteria, A is the initial, t is the time.

We want to solve for t when G = 700 and A = 4.

We substitute into the equation and get
`700=4e^(0.6t)`, which is
`(700)/(4)=e^(0.6t)`.

To solve for t, we need to remember that the natural logarithm is the inverse of an exponent with base e. We take the ln of both sides like so.


`ln( (700)/(4))=ln(e^(0.6t))\\\\ln( (700)/(4))=0.6t\\\\t= (5)/(3) ln( (700)/(4)) = 8.608 \ hours`

8) We are solving the equation for t when P = 5,000, r = 2% = 0.02, n = 12 (it's compounded 12 times in a year), and A = 10,000.


`10000=5000(1+ (0.02)/(12))^(12t)`

We first divide both sides by 5,000. Then, we take the natural log (ln) of both sides and simplify to solve for t.

t = 34.686 years