Question

A population P is initially 1000. How do you find an exponential model (growth or decay) for the population after t years if the population P is increases by 200% every 6 years?

Answer 1

This question is from the topic of pre-calculus. In solving this question, we will first write the formula for this exponential model which is increasing with time. Here, we will multiply a constant ‘k’ with time t for knowing the growth rate. After that, we will solve the formula of the exponential model using the formulas of logarithms. After solving the further question, we will get our answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have given that the number of population initially is 1000. And, it is saying that it is increasing by 200% after every 6 years. So, we have to find an exponential model for the population after t years.
Let us suppose P as population after t year and ${{\text{P}}_{0}}$ as initial population that is 1000 and take t in years. The general formula for exponential model can be written as
$P={{P}_{0}}{{e}^{kt}}$
Where, k is growing rate constant. Let us find out the value of k.
The above equation can also be written as
$\Rightarrow \dfrac{P}{{{P}_{0}}}={{e}^{kt}}$
In the above equation, let us take $\ln$ (log base e or we can say ${{\log }_{e}}$) to both side of equation, we get
$\Rightarrow \ln \left( \dfrac{P}{{{P}_{0}}} \right)=\ln \left( {{e}^{kt}} \right)$
In the question, we have seen that after 6 years the population is increasing by 200%.
So, after 6 years, the population will be
$P={{P}_{0}}+\left( 200 \right)\dfrac{{{P}_{0}}}{100}={{P}_{0}}+2{{P}_{0}}=3{{P}_{0}}$
Hence, the new population after years will be 3 times the initial population in 6 years. So, we can write
$\Rightarrow \ln \left( \dfrac{3{{P}_{0}}}{{{P}_{0}}} \right)=\ln \left( {{e}^{k\times 6}} \right)$
$\Rightarrow \ln 3=\ln \left( {{e}^{6k}} \right)$
Now, using the formula of logarithms that is $\ln \left( {{x}^{a}} \right)=a\ln x$, we can write
$\Rightarrow \ln 3=6k\ln e$
Now, using the formula of logarithms that is $\ln e=1$, we can write
$\Rightarrow \ln 3=6k$
$\Rightarrow 6k=\ln 3$
The above equation can also be written as
$\Rightarrow k=\dfrac{1}{6}\ln 3$
Using the formula $\ln \left( {{x}^{a}} \right)=a\ln x$, we can write
$\Rightarrow k=\ln {{3}^{\dfrac{1}{6}}}$
Hence, we have found the growth rate. Then, exponential model will be
$P={{P}_{0}}{{e}^{t\times \ln {{3}^{\dfrac{1}{6}}}}}$
This can also be written as by using the formula $\ln \left( {{x}^{a}} \right)=a\ln x$,
$\Rightarrow P={{P}_{0}}{{e}^{\ln {{3}^{\dfrac{t}{6}}}}}$
Now, using the formula ${{e}^{\ln {{a}^{b}}}}={{a}^{b}}$, we can write
$\Rightarrow P={{P}_{0}}\left( {{3}^{\dfrac{t}{6}}} \right)$
Hence, we have found the exponential model. The model is $P={{P}_{0}}\left( {{3}^{\dfrac{t}{6}}} \right)$

Note: For solving this type of question easily, we should have a better knowledge in the topic of pre-calculus. We should remember the formulas for solving this type of question easily:
${{e}^{\ln {{a}^{b}}}}={{a}^{b}}$
$\ln e=1$
$\ln \left( {{x}^{a}} \right)=a\ln x$
Remember that if it is growth, then the exponential model will be like $P={{P}_{0}}{{e}^{kt}}$ and if it is decay, then exponential model will be like $P={{P}_{0}}{{e}^{-kt}}$.