# Suppose that the population of a colony of bacteria increases exponentially. If the population at the start is 300 and 4 hours later it is 1800, how long (from the start) will it take for the population to reach 3000?

## A) $\text{4 ln10"/"ln6}$ $\text{hrs}$ B) "6/(4ln10) $\text{hrs}$ C) "(10ln4)/ln6 $\text{hrs}$ D) "4(ln10-ln6) $\text{hrs}$ Answer is A). Thanks in advance!

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#### Explanation

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#### Explanation:

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Jun 12, 2018

See below.

#### Explanation:

We need get an equation of the form:

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

Where:

$A \left(t\right)$ is the amounf after time t ( hours in this case ).

$A \left(0\right)$ is the starting amount.

$k$ is the growth/decay factor.

$t$ is time.

We are given:

$A \left(0\right) = 300$

$A \left(4\right) = 1800$ ie after 4 hours.

We need to find the growth/decay factor:

$1800 = 300 {e}^{4 k}$

Divide by 300:

${e}^{4 k} = 6$

Taking natural logarithms of both sides:

$4 k = \ln \left(6\right)$ ( $\ln \left(e\right) = 1$ logarithm of the base is always 1 )

Divide by 4:

$k = \ln \frac{6}{4}$

Time for population to reach 3000:

$3000 = 300 {e}^{\frac{t \ln \left(6\right)}{4}}$

Divide by 300:

${e}^{\frac{t \ln \left(6\right)}{4}} = 10$

Taking logarithms of both sides:

$\frac{t \ln \left(6\right)}{4} = \ln \left(10\right)$

Multiply by 4:

$t \ln \left(6\right) = 4 \ln \left(10\right)$

Divide by $\ln \left(6\right)$

t=color(blue)((4ln(10))/(ln(6)) " hrs"

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