On June $3$ at $10$ am, I find out $2 $ % of the surface of the fish tank is covered with moss. On june $8$ at $10$ am, I see that my fish tank now has $4$ % of the surface covered with moss.
$Qs:$ Write $F(t)$ to represent the amount of moss covering the surface of the fish tank at time $t$ ($t$ is measured in days starting June $3$ at $10$ am) if the growth is exponential. And write $L(t)$ if the growth is linear.
Attempt:
Since growth is exponential, I want to find $F(t) = cb^{kt}$. I know $F(0) = 0.02$. So $F(t) = 0.02b^{kt}$. Also, $F(5) = 0.02b^{5k} = 0.04$. This gives $b^{5k} = 2 $ and so $5k = \log_b 2 $. Hence, $k = \log_b 2 $.
To find $L(t)$. We know $L(0) = 0.02$ and $L(5) = 0.04$ and so the slope of $L$ is $ \frac{ .04-.02}{5} = \frac{ .02}{5} = \frac{2}{500} = \frac{1}{250}$. and so the line is given by $L(t) - 0.02 = \frac{t}{250} $
IS this a correct solution? I am still unsure on the first part on how can we find $b$.
You may simplify your function $F$ by just writing $$ F(t) = cb^{t}. $$ No need to bring another parameter $k$.
Then you write $b^5=2$, giving $b=2^{1/5}$.