# Assume that the speed of the new computers costing a certain amount of money grows exponentially...

## Question:

Assume that the speed of the new computers costing a certain amount of money grows exponentially over time, with the speed doubling every 18 months. Suppose that you need to run a very time-consuming program and you are allowed to buy a new computer costing a fixed amount to run the program. If you start now, the program will take 6 years to finish. How long should you delay buying the computer and starting the program so that it will finish in the shortest amount of time from now? Also, what will the shortest finishing time be?

## Exponential Growth

This problem is based on the concept of exponential growth. We are going to use the formula to make some substitutions and based on that we are going to calculate the delay and the shortest time.

Let the speed of the new computer after {eq}t {/eq} months be,

{eq}\displaystyle S(t)=S_0e^{kt} {/eq}

It is given that the speed will double in 18 months,

{eq}\displaystyle 2S_0=S_0e^{k\times 18} {/eq}

{eq}\displaystyle 2=e^{k\times 18} {/eq}

Taking logarithms on both the sides, we get,

{eq}\displaystyle k=\frac{1}{18}\ln (2)=0.04 {/eq}

It is also given that it will take 6 years to finish the program. So, in 6 years the speed will be,

{eq}\displaystyle \frac{6 \times 12}{18}=4 {/eq} times.

So,

{eq}\displaystyle 2^4S_0=S_0e^{0.04t} {/eq}

{eq}\displaystyle 16S_0=S_0e^{0.04t} {/eq}

{eq}\displaystyle 16=e^{0.04t} {/eq}

Taking logarithm on both the sides, we get,

{eq}\displaystyle \ln 16=0.04t \ln e {/eq}

{eq}\displaystyle 2.7725=0.04t {/eq}

Thus,

{eq}\displaystyle \boxed{\displaystyle t \approx 69 \text{ months}} {/eq}

• {eq}\displaystyle \text{So, the delay in buying the computer should be about 69 months. } {/eq}
• And thus, the shortest amount of time should be, {eq}72-69=3 \text{ months} {/eq}. 