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}