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7) We are given 
. 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
, which is 
.
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.
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.
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
We want to solve for t when G = 700 and A = 4.
We substitute into the equation and get
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.
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.
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