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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...

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

  • 0
    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.2012-01-12

3 Answers 3

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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.

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    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.2015-04-26
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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.

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    @ArturoMagidin, There, I added it2012-01-12
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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.

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    Alright, I could have done something like that. Thanks.2012-01-12