Question

A country’s population in 1995 was 164 million. In 2001 it was 169 million. How do you estimate the population in 2015 using an exponential growth formula?

Answer 1

Now we know that the exponential growth function is given by $f\left( x \right)=a{{\left( 1+\dfrac{r}{100} \right)}^{t}}$ where a is the initial value r is the rate of growth and t is the time in years. Now we have In 1995 was 164 million and in 2001 it was 169 million. Hence we will use this condition and find the value of $\left( 1+\dfrac{r}{100} \right)$ . Now we will again use the formula $f\left( x \right)=a{{\left( 1+\dfrac{r}{100} \right)}^{t}}$ to find the population in 2015 years by substituting corresponding value of t.

Complete step-by-step solution:
Now in 1995 the population was 164 million.
In 2001 the population was 169 million.
Now the time span between the two is 2001 – 1995 = 6 years.
Hence we have the time span is 6 years.
Now let us see the exponent growth formula $f\left( x \right)=a{{\left( 1+r \right)}^{t}}$
Now here a is the initial value r is the rate of growth and t is the timespan.
Now the population after 6 years was 169 billion and initial population was 164 billion.
Hence we have a = 164 billion, t =6 years and $f\left( x \right)=169$ billions.
Hence we get,
$\begin{aligned} & \Rightarrow 169=164{{\left( 1+\dfrac{r}{100} \right)}^{6}} \\\ & \Rightarrow \dfrac{169}{164}={{\left( \dfrac{100+r}{100} \right)}^{6}} \\\ \end{aligned}$
Now taking log on both sides we have,
$\Rightarrow \log \left( \dfrac{169}{164} \right)=\log {{\left( \dfrac{100+x}{100} \right)}^{6}}$
Now we know ${{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x$ Hence using this we get,
$\Rightarrow \log \left( \dfrac{169}{164} \right)=6\left[ \log \left( \dfrac{100+r}{100} \right) \right]$
Now again we have the property of log which states $\log \dfrac{a}{b}=\log a-\log b$ . Hence using this we get,
$\Rightarrow \log 169-\log 164=6\left[ \log \left( \dfrac{100+r}{100} \right) \right]$
Now substituting the values we get,
$\begin{aligned} & \Rightarrow 2.228-2.215=6\left[ \log \left( \dfrac{100+r}{100} \right) \right] \\\ & \Rightarrow 0.013=6\left[ \log \left( \dfrac{100+r}{100} \right) \right] \\\ & \Rightarrow \left[ \log \left( \dfrac{100+r}{100} \right) \right]=\dfrac{0.013}{6} \\\ & \Rightarrow \left( \dfrac{100+r}{100} \right)={{10}^{\dfrac{0.013}{6}}} \\\ \end{aligned}$
Now let us say we want to find the population in 2015.
Now in 2001 the population was 169 million.
Now 2015 – 2001 = 14.
Hence the time span is 14 years.
Now the population after 14 years is, $169{{\left( 1+\dfrac{r}{100} \right)}^{14}}$
$\Rightarrow 169\times {{10}^{\dfrac{0.013\times 14}{6}}}$

& \Rightarrow 169\times {{10}^{0.03}} \\\ & \Rightarrow 169\times 1.07 \\\ & \Rightarrow 181 \\\ \end{aligned}$$ **Hence the population in 2015 will be 181 million approximately.** **Note:** Now note that in the term ${{\left( 1+\dfrac{r}{100} \right)}^{t}}$ r is the rate of growth in percentage. Hence we have the factor divided by 100. Also note that we can use log or ln for calculation as per convenience.