# Thread: How I can solve this question

1. ## How I can solve this question

Hiu

How I can solve this question

2. Originally Posted by r-soy
Hiu

How I can solve this question
To graph the equation between 0 and 5 start with the ordered point (0,100)

You know that in 1 hour the amount will be $2^2 = 4$ times the amount at 0 so you get the second ordered pair of (1,400)

Plot the others similarly and draw the graph

Spoiler:
Use the exponential growth equation $A_t = A_0e^{kt}$

We know that $A_0 = 100$ and that in $t = \frac{1}{2}$ the number doubles - $A_t = 2A_0 = 200$

Use this relationship to find the growth constant $k$

$200=100e^{\frac{1}{2}k}$

$k = 2ln(2) = ln(4)$

Spoiler:
$A_t=A_0e^{t\,ln(4)} = A_0 \cdot 4^t$

As the graph is exponential growth I suggest using a log scale on the y axis

$ln(y) = kt + ln(A_0)$ which gives a straight line in the form $y=mx+c$

Edit: Plot $ln(y)$ against $t$ to get the straight line, with gradient $k$ and y intercept $ln(A_0)$

3. thanks e^(i*pi)

4. There is no reason to use "e" or logarithms.

There are 2N half hour periods in t hours so the population will have doubled 2t times: $N= (100) (2^{2t})$