The number, N, of people who have heard a rumor spread by mass media by time, t, is given by...

Question:

The number, N, of people who have heard a rumor spread by mass media by time, t, is given by {eq}N\left( t \right)=a\left( 1-{{e}^{-kt}} \right) {/eq}. There are 6 million people in the population, who hear the rumor eventually. If 5% of them heard it on the first day, find the percentage of the population who have heard the rumor after 5 days.

Growth Equation

Exponential growth and decay equations are equations that takes an initial population and describe the increase of population as time progress. The problem is similar to growth equations, where we substitute the initial population and desired time to solve for the population at that time.

Answer and Explanation:

The rumor was spread to the {eq}5 \% {/eq} of the population on the first day. In the equation {eq}N\left(t\right)=a\left(1-e^{-k\left(t\right)}\right) {/eq}, {eq}a {/eq} refers to the initial number of people who heard the rumor before it starts spreading. Then writing into the equation, the spread of rumor on the first day

$$\begin{align} N\left(t\right)&=a\left(1-e^{-k\left(t\right)}\right) \\ 0.05a&=a\left(1-e^{-k\left(1\right)}\right) \\ 0.05&=1-e^{-k\left(1\right)} \\ e^{-k\left(1\right)}&=1-0.05 \\ e^{-k\left(1\right)}&=0.95 \end{align} $$

To simplify this, we can eliminate Euler's number by taking the natural logarithms of both sides of the equation

$$\begin{align} ln\left(e^{-k\left(1\right)}\right)&=\ln\left(0.95\right) \\ -k\left(1\right)&=\ln\left(0.95\right) \\ k&=-\ln\left(0.95\right) \\ \end{align} $$

By substitution, we can now write the function for the spread of rumor as follows.

$$\begin{align} N\left(t\right)&=a\left(1-e^{-\left(-ln\left(0.95\right)\right)\left(t\right)}\right) \\ N\left(t\right)&=a\left(1-e^{ln\left(0.95\right)\left(t\right)}\right) \\ N\left(t\right)&=a\left(1-e^{tln\left(0.95\right)}\right) \\ N\left(t\right)&=a\left(1-e^{ln\left(0.95\right)^t}\right) \\ N\left(t\right)&=a\left(1-\left(0.95\right)^t\right) \\ \end{align} $$

Now that we have the required function, we can now solve for {eq}N(t) {/eq} at the instant when {eq}t = 5 {/eq}

$$N\left(5\right)=a\left(1-\left(0.95\right)^{\left(5\right)}\right) \\ N\left(5\right)=a\left(0.2262190625\right) \\ $$

Then solving for the percentage

$$\begin{align} \%\text{inPopulation}&=\frac{N\left(5\right)}{a}\times 100\% \\ \%\text{inPopulation}&=\frac{0.2262190625a}{a}\times 100\% \\ \%inPopulation&=22.62\% \end{align} $$

Therefore, 22.62% of the total population have heard the rumor in 5 days.


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Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

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