I have the following math problem:
The number of people in a town of 10,000 who have heard a rumor started by a small group of people is given by the following function: $N(t) = \frac{10000}{5+1245e^{-0.97t}}$
As far as I can get without knowing e is: $\frac{10000}{5+1245e^{-4.85}}$.
Trying to use logarithms I get $-4.85 = \ln{\frac{2000-x}{249x}}$, which seems to be a dead end.
I'm in an online precalculus course and they made no mention of the value of the natural number, e, nor how to solve equations that use it. Am I missing something, or is it impossible to solve this without using the value of $e$?
edit: sorry, forgot to add that the question for the problem is:
On day 5, approximately how many people had heard the rumor?
edit: How I got $-4.85 = \ln{\frac{2000-x}{249x}}$ is:
$N(t) = \frac{10000}{5+1245e^{-0.97t}}$
so
$N(5) = \frac{10000}{5+1245e^{-0.97*5}}$
Which is
$N(5) = \frac{10000}{5+1245e^{-4.85}}$
Solving for $N(5)$ as $x$
$x = \frac{10000}{5+1245e^{-4.85}}$
Multiplying both sides by $5+1245e^{-4.85}$
$(x)(5+1245e^{-4.85})=10000 $
Dividing both sides by $x$
$5+1245e^{-4.85}=\frac{10000}{x} $
Subtracting $5$ from both sides
$1245e^{-4.85}=\frac{10000}{x}-5 $
Dividing both sides by $1245$
$e^{-4.85}=\frac{10000}{1245x}-\frac{5}{1245} $
Taking natural log of both sides
$\ln{e^{-4.85}}=\ln{\frac{10000}{1245x}-\frac{5}{1245} }$
Simplifying natural log and fractions on the right
$-4.85=\ln{\frac{2000}{249x}-\frac{1}{249} }$
Getting common denominator on the right
$-4.85=\ln{\frac{2000}{249x}-\frac{x}{249x} }$
And...
$-4.85=\ln{\frac{2000-x}{249x} }$