# A certain bacteria population P obeys the exponential growth law P(t) = 3500e^{0.6t}, where t is...

## Question:

A certain bacteria population {eq}P {/eq} obeys the exponential growth law {eq}P(t) = 3500e^{0.6t} {/eq}, where {eq}t {/eq} is the time in hours beginning at {eq}t=0 {/eq}.

A) How many bacteria are present initially?

B) At what time will there be 10,000 bacteria?

## Exponential Growth:

Suppose that {eq}Q(t) {/eq} is a quantity which varies with time. We say that {eq}Q(t) {/eq} grows exponentially if {eq}Q(t)=Q_0e^{kt} {/eq} for some positive constants {eq}Q_0 {/eq} and {eq}k {/eq}. The constant {eq}Q_0 {/eq} is called the initial value of {eq}Q(t) {/eq}, while the constant {eq}k {/eq} is called the growth constant of {eq}Q(t) {/eq}.

A) The number of bacteria present initially is given by {eq}P(0)=3500e^{0.6 \cdot 0}=3500 {/eq}. So there were {eq}\boxed{3500} {/eq} bacteria present initially.

B) In order to find when there will be 10,000 bacteria present, we must solve the equation {eq}P(t)=10000 {/eq}. We do so as follows:

{eq}\begin{align*} 10000&=P(t)\\ &=3500e^{0.6t}\\ \frac{10000}{3500}&=e^{0.6t}\\ \frac{20}{7}&=e^{0.6t}\\ \ln \frac{20}{7}&=0.6t&\text{(taking the logarithm of both sides)}\\ \frac{1}{0.6}\ln \frac{20}{7}&=t&\text{(isolating }t\text{).} \end{align*} {/eq}

So {eq}t=\frac{1}{0.6}\ln \frac{20}{7} {/eq}. Since {eq}t {/eq} is measured in hours, this means that there will be 10,000 bacteria after {eq}\boxed{\frac{1}{0.6}\ln \frac{20}{7}\text{ hr} \approx 1.75\text{ hr}}\, {/eq}.