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:

$N(t) = \frac{10000}{5+1245e^{-0.97t}}$


$N(5) = \frac{10000}{5+1245e^{-0.97*5}}$

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

$-4.85=\ln{\frac{2000}{249x}-\frac{x}{249x} }$


$-4.85=\ln{\frac{2000-x}{249x} }$

  • $\begingroup$ You didn't state the problem. I am guessing it is something like "How many have heard the rumour after $5$ days?" (or maybe hours, units were not mentioned). You need $e^{-4.85}$. Any scientific calculator will find it for you, the button to press is the $e^x$ button. An approximate calculation by hand is possible, but very unpleasant. $\endgroup$ – André Nicolas Jan 12 '12 at 4:39
  • $\begingroup$ If the question is interpreted as meaning how many have heard on day $5$ but not earlier, need $N(5)-N(4)$. $\endgroup$ – André Nicolas Jan 12 '12 at 4:45
  • $\begingroup$ 'e' isn't a natural number as you say. It is transcendental proof of which is a highly non-trivial one! $\endgroup$ – user21436 Jan 12 '12 at 8:07
  • $\begingroup$ And what exactly were you hoping to do with your manipulations of the equation? You now have a logarithmic equation in $x$; if you want to find the value of $x$, you are going to have to undo everything you did and re-introduce the exponential function with base $e$. Simply put: to find an approximate value of $x$, you need an approximate value of $e$. There is no way around it. $\endgroup$ – Arturo Magidin Jan 12 '12 at 17:56

Note that if $e^x=10^y$ then taking natural logarithms $x=y \ln(10)$ so $y=x/\ln(10)$. For what it is worth, $\ln(10)\approx 2.302585$.

So you can rewrite your formula as $$N(t) = \frac{10000}{5+1245 \times 10^{-0.97t/\ln(10)}} \approx \frac{10000}{5+1245 \times 10^{-0.421t}}.$$

I don't know if this helps: I would have thought that if you are willing to use natural logarithms, you should be willing to use their inverses.

  • $\begingroup$ Alright, I could have done something like that. Thanks. $\endgroup$ – mowwwalker Jan 12 '12 at 19:27

If you are uncertain what the constant $e$ is I suggest you look at the following website: http://en.wikipedia.org/wiki/E_%28mathematical_constant%29

A good approximation for a simple question like this would be: $$e \approx 2.72$$

The reason the formula given to you for that kind of "growth rate" problem is that $$x(t)=ae^{kt}$$ is the general exponential function where $a$ and $k$ are constants that relate to your specific problem.

  • $\begingroup$ I know what e is and it's approximate value, but I'm saying we were never taught it, so I'm wondering if it's possible to solve the equation without it... If you don't know the value of e, is it still possible to solve that equation? $\endgroup$ – mowwwalker Jan 12 '12 at 4:23
  • 2
    $\begingroup$ @Walkerneo: You can leave $e$ indicated. If your question is whether one can deduce the value of $e$ from these equations and this problem, the answer is "no, you cannot." $\endgroup$ – Arturo Magidin Jan 12 '12 at 4:35
  • $\begingroup$ My question is if I can deduce the number of people, which I think ended up being 678, from that equation without using the value of e. $\endgroup$ – mowwwalker Jan 12 '12 at 4:38
  • $\begingroup$ Care to write out your steps for how you arrived at $-4.85 = \ln(\frac{2000-x}{249x})$? $\endgroup$ – Samuel Reid Jan 12 '12 at 4:38
  • $\begingroup$ @SamuelReid, Sorry, I forgot to mention that the question asks how many people had heard to rumor on day 5. $\endgroup$ – mowwwalker Jan 12 '12 at 4:41

I am afraid you do need to use the constant $e$ and that you did not get rid of it when you took the natural logarithm of both sides, you just hid it in the notation. The reason why this is true is because $\ln f(x)=\log_{e}{f(x)}$ and what this basically means is:

What do I have to take to the power of $e$ to get the value of $f(x)$

In any case, if you are curious as to what the value of $e$ is, it is defined by the following limit: $\lim_{n\rightarrow \infty}{\left(1+\frac{1}{n}\right)^n}\approx 2.71828$.

For further reading you can read this Wikipedia article.

  • $\begingroup$ For $n_1 \lt n_2$ it also holds that $(1+\frac{1}{n_1})^{n_1} \lt (1+\frac{1}{n_2})^{n_2} \lt e \lt (1+\frac{1}{n_2})^{n_2+1} \lt (1+\frac{1}{n_1})^{n_1+1}$ which is all one needs to know in order to approximate $e$ arbitrary well. $\endgroup$ – Keba Apr 26 '15 at 21:48

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