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

7) We are given [tex]G=Ae^{0.6t}[/tex]. 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 [tex]700=4e^{0.6t}[/tex], which is [tex] \dfrac{700}{4}=e^{0.6t}[/tex].

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.

[tex]ln( \frac{700}{4})=ln(e^{0.6t})\\\\ln( \frac{700}{4})=0.6t\\\\t= \frac{5}{3} ln( \frac{700}{4}) = 8.608 \ hours[/tex]

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. 

[tex]10000=5000(1+ \dfrac{0.02}{12})^{12t}[/tex]

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