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I’m not sure how to do this, I believe it is exponential functions.

I’m not sure how to do this, I believe it is exponential functions.-example-1
User Rahpuser
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Exponential Growth

The growth of a given variable y with an initial value of Po can be modeled with an exponential function as follows:


y=P_o(1+r)^t

Where r is the growth rate and t is the time.

In 2006 (here we assume t = 0), the number of cell phone subscribers was Po = 233 million. We know the number of subscribers increases by r = 6% every year.

a.

With the given information, we can write the exponential function. We must express the rate in decimal, thus r = 6/100 = 0.06.

Our model is expressed as:


\begin{gathered} y=233(1+0.06)^t \\ Operate\colon \\ y=233(1.06)^t \end{gathered}

This is the exponential model.

The number of subscribers in 2008 can be estimated with our model. The value of t is t = 2008 - 2006 = 2 years:


\begin{gathered} y=233(1.06)^2 \\ y=233\cdot1.1236 \\ y=262 \end{gathered}

The estimated number of subscribers in 2008 is 262 million.

b.

It's required to find the value of t when y = 278. Substituting in the model function:


278=233(1.06)^t

We need to solve for t. Dividing by 233:


(278)/(233)=1.06^t

Taking natural logarithms on both sides:


\begin{gathered} \ln (278)/(233)=\ln 1.06^t \\ \ln (278)/(233)=t\ln 1.06 \\ t=(\ln (278)/(233))/(\ln 1.06) \end{gathered}

Calculating:

t = 3 years.

In 2009, the number of cell phone subscribers was about 278 million

User Vjardel
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