# The number of mold spores in a petri dish increases by a factor of 10 every week. If there are initially 40 spores in the dish, how long will it take for there to be 2000 spores?

Dec 19, 2017

$0.8495$ weeks

5 days, 22 hours ( to the nearest hour )

#### Explanation:

We need to find an equation of the form:

$A \left(t\right) = {A}_{0} {e}^{k t}$

Where ${A}_{0}$ is the initial amount, $k$ is the growth/decay factor, $A \left(t\right)$ is the amount after time $t$ and $t$ is the time. For this example we will take $t$ to be in weeks.

From the given information we know the initial amount is 40.

If they are increasing by a factor of 10 every week, then we would expect the amount after 1 week to be 400. So, using our equation:

$400 = 40 {e}^{k}$ ( t = 1 for 1 week)

We need to solve this to find the growth/decay factor $k$.

Divide both sides by 40:

$100 = {e}^{k}$

Taking natural logs of both sides:

$\ln \left(100\right) = k \ln \left(e\right)$ ( ln(e)=1, the logarithm of the base is always 1)

$k = \ln \left(100\right)$

Now we know $k$ we can solve the problem for $t$:

Final amount is $2000$, So:

$2000 = 40 {e}^{\ln \left(100\right) t}$

Using the fact that ${e}^{\ln} \left(a\right) = a$

$2000 = 40 {\left(100\right)}^{t}$

Divide by 40:

$50 = {\left(100\right)}^{t}$

Taking logs of both sides:

$\ln \frac{50}{\ln} \left(100\right) = t \implies t = 0.8495$ weeks (4 .d.p)

or 5 days, 22 hours ( to the nearest hour )

CHECK:

$A \left(t\right) = 40 {e}^{\ln \left(100\right) t}$

$A \left(t\right) = 40 {e}^{\ln \left(100\right) \cdot 0.8495}$

$A \left(t\right) = 40 \cdot {\left(100\right)}^{0.8495} = 2000.13814$

We wouldn't expect this to be exact, because we rounded to 4 dp.