# A recent study shows that the number of Australian homes with a computer doubles every 8 months....

## Question:

A recent study shows that the number of Australian homes with a computer doubles every 8 months. Assuming that the number is increasing continuously, at approximately what monthly rate must the number of Australian computer owners be increasing for this to be true?

## Exponential growth

The continuous growth on a number that is a part of the increasing number. For example, if {eq}100 {/eq} is increasing at a rate of {eq}10\% {/eq} after each day.

The number will increase by {eq}10\% {/eq} of itself.

After 1 day, the value will become {eq}100 + 10\%\ of\ 100 = 110 {/eq}.

After 2 days, the value will become {eq}110 + 10\%\ of\ 110 = 111 {/eq} and so on.

The value is increasing in the following pattern for {eq}T {/eq}.: $$\text{Final value= Initial value}\times \dfrac{Rate}{100} \times \dfrac{Rate}{100} \times \dfrac{Rate}{100}....... T\ times$$

$$\text{Final value= Initial value}\times (1 + \dfrac{Rate}{100})^{T}$$

Let us assume that the number of Australian homes with a computer is {eq}a {/eq} and increasing the rate of {eq}r\% {/eq} per month and the number increases to {eq}2a {/eq} after {eq}8\ monts. {/eq}

Using exponential growth formula:

$$\text{Final value= Initial value}\times (1 + \dfrac{Rate}{100})^{Time}$$

$$2a=a\times (1+\dfrac{r}{100})^8$$

Dividing the equation by {eq}a: {/eq}

$$2=(1+\dfrac{r}{100})^8$$

Taking log on both sides

$$log(2)= 8\times log(1+\dfrac{r}{100})$$

$$0.3010 = 8\times log(1+\dfrac{r}{100})$$

Dividing the equation by 8

$$0.0376 = log(1+\dfrac{r}{100})$$

$$1.0904 = 1+\dfrac{r}{100}$$

$$\dfrac{r}{100} = 1.0904 - 1$$

$$\dfrac{r}{100} = 0.0904$$

Multiplying theequation by 100

$$r= 0.0904*100$$

$$r= 9.04\%$$

The monthly rate of growth is {eq}9.04\%. {/eq} 