Question

How do you find the associated exponential growth model: Q = 3000 when t = 0 and doubling time = 3.

Answer 1

We are given Q = 3000 at t = 0 and doubling time is 3. We have to find the exponential growth function. To do so we will learn how exponential functions are defined. Then we will learn about what doubling time is. Using the time given to us as 3, we will try to find the missing constant. We will be using $Q=C{{e}^{kt}}$ where C and k are constants.

Complete step-by-step solution:
We are asked to find the exponential growth model. To do so we will learn about the exponential model in one of the models that helps us to tell about the growth in decay (decrease) of any function (condition). Generally, an exponential function is given us as
$Q=C{{e}^{kt}}$
where C tells us about the initial value of the function and k is constant. Now, we will use our given details to fill up to the constant and will form our exponential function for our growth model. Now, we have given that when t = 0, we have Q = 3000. We will put t = 0 in $Q=C{{e}^{kt}}$ and use Q = 3000. So, we will get,
$\Rightarrow 3000=C{{e}^{k\times 0}}$
As $k\times 0=0$ so we get,
$\Rightarrow 3000=C{{e}^{0}}$
As we know ${{e}^{0}}=1,$ so we get,
$\Rightarrow 3000=C$
So, we get C = 3000. Now we have the doubling time as 3. The doubling time means that the time required for the value of the function to double from the initial amount. This means we have time t = 3 in which Q will be 6000 from 3000. So, we use Q = 6000, C = 3000, t = 3 in $Q=C{{e}^{kt}}$ and solve for k. So, using the above value, we get,
$6000=3000{{e}^{k\times 3}}$
On simplifying, we get,
$\Rightarrow 6000=3000{{e}^{3k}}$
On solving this, we get,
$\Rightarrow 2={{e}^{3k}}$
Taking log on both the sides, we get,
$\Rightarrow \ln 2=\ln \left( {{e}^{3k}} \right)$
As $\ln \left( {{a}^{b}} \right)=b\ln \left( a \right)$ so we get,
$\Rightarrow \ln 2=3k\ln \left( e \right)$
As ln (e) = 1, so we get,
$\Rightarrow \ln 2=3k$
Dividing both the sides by 3, we get,
$\Rightarrow k=\dfrac{\ln \left( 2 \right)}{3}$
Hence, we get, C = 3000 and $k=\dfrac{\ln \left( 2 \right)}{3}.$ Using these in $Q=C{{e}^{kt}}$ we get,
$\Rightarrow Q=3000{{e}^{\dfrac{\ln \left( 2 \right)}{3}t}}$
So our required exponential function is $Q=3000{{e}^{\dfrac{\ln \left( 2 \right)}{3}t}}.$

Note: Remember in the exponential function, the value of the function increases more rapidly when we put a higher value, the graph of the exponential function reaches infinity for the higher value. In linear function, the graph is more by some difference but in the exponent, the one value may vary too much from another value, so we need to calculate carefully.