# How do I solve this exponential equation problem?

May 17, 2018

#### Explanation:

The first question asks for $P \left(13\right)$:

$P \left(13\right) = 147 {\left(1 + 0.0151\right)}^{13}$

$P \left(13\right) \approx 178.62 \text{ million}$

The second question asks for $P \left(20\right)$

$P \left(20\right) = 147 {\left(1 + 0.0151\right)}^{20}$

$P \left(20\right) \approx 198.38 \text{ million}$

For the third question, the population equal to 2 billion translates into setting $P \left(t\right) = 2000$ and then solving for t:

$2000 = 147 {\left(1 + 0.0151\right)}^{t}$

$\frac{2000}{147} = {\left(1 + 0.0151\right)}^{t}$

$\ln \left({\left(1 + 0.0151\right)}^{t}\right) = \frac{2000}{147}$

$t = \frac{2000}{147 \ln \left(1 + 0.0151\right)}$

$t = 907.8$

The year will be $2907$.

May 17, 2018

1) $178.621 \text{ million people}$
2) $198.378 \text{ million people}$
3) The year will be 2052.

#### Explanation:

To emphasize, $P \left(t\right)$ measures population in millions, and t represents the number of years since 2000.

$P \left(t\right) = 147 {\left(1 + .0151\right)}^{t}$
$P \left(t\right) = 147 {\left(1.0151\right)}^{t}$

Plug in values:

1) The problem asks about the year 2013, which is $\textcolor{b l u e}{13}$ years after the year 2000.

$P \left(\textcolor{b l u e}{13}\right) = 147 {\left(1.0151\right)}^{\textcolor{b l u e}{13}}$

$P \left(13\right) = 178.620827 \text{ million people}$

2) The year 2020 is $\textcolor{b l u e}{20}$ years after the year 2000.

$P \left(\textcolor{b l u e}{20}\right) = 147 {\left(1.0151\right)}^{\textcolor{b l u e}{20}}$

$P \left(20\right) = 198.378175 \text{ million people}$

3) Set 2 billion people equal to $P \left(t\right)$. Recall that $P \left(t\right)$ measures in millions. $\textcolor{g r e e n}{2 \text{ billion" = 2,000" million}}$

$2 , 000 = 147 {\left(1.051\right)}^{t}$

${1.051}^{t} = \frac{2 , 000}{147}$

$t = {\log}_{1.051} \left(\frac{2 , 000}{147}\right)$

$t = 52.48$

$\text{Year } = 2000 + 52.48 = 2052$

The year will be 2052.

Note: If you have only a scientific calculator that can perform logarithms on only base $e$ or base $10$, use the logarithm change of base formula ${\log}_{b} \left(x\right) = {\log}_{a} \frac{x}{\log} _ a \left(b\right)$, so

${\log}_{1.051} \left(\frac{2 , 000}{147}\right) = \frac{{\log}_{a} \left(\frac{2000}{147}\right)}{{\log}_{a} \left(1.051\right)}$

where $a$ is whatever base you choose.