We need an expression for the population at time *t*.

After one year, the population will be `1.3%` higher than in 2000. (1.3% = 0.013)

Population after 1 year: `6\ "billion" × 1.013`.

Population after 2 years: `6\ "billion" × (1.013)^2`.

Population after 3 years: `6\ "billion" × (1.013)^3`.

So our population, *P*, after *t* years, is given by:

`P(t) = 6\ "billion" × (1.013)^t`

[In general, for any population growth,

`P(t) = P_0(1 + r)^t`

where *P*_{0} is the
population at time `t = 0`, *r* is the rate of growth per time period and *t* is the time.]

We are asked to find when the population doubles, so we need to solve:

`12 000 000 000 =` ` 6 000 000 000 × (1.013)^t`

This gives `2 = (1.013)^t`

Taking logarithms of both sides, we have:

`log\ 2 = log (1.013)^t`

Using the third log law, we have:

`log\ 2 = t\ log\ 1.013`

So

`t=(log\ 2)/(log\ 1.013)=53.66`

So it will take only about `54` years to double the world's population, if it continues to grow at the current rate.

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