Question

1. if nothing is done and the bacteria continue to double, how long will it take for the count to reach 3 million bacteria? write an equation to represent this situation. find at least 2 different ways to solve the equation.

Explanation:

Step1: Assume initial bacteria count and time - step

Let the initial bacteria count $N_0 = 1$ (for simplicity, since we are interested in the growth factor) and assume the bacteria double every unit - time $t$. The general formula for exponential growth of bacteria is $N = N_0\times2^t$, where $N$ is the final number of bacteria and $t$ is the number of time - steps. We want $N = 3\times10^6$, so the equation is $3\times10^6=1\times2^t$.

Step2: Solve using logarithms

Take the natural logarithm of both sides of the equation $3\times10^6 = 2^t$. We know that $\ln(a^b)=b\ln(a)$, so $\ln(3\times10^6)=t\ln(2)$.
First, use the property $\ln(ab)=\ln(a)+\ln(b)$: $\ln(3) + \ln(10^6)=t\ln(2)$. Since $\ln(10^6)=6\ln(10)\approx6\times2.3026 = 13.8156$ and $\ln(3)\approx1.0986$, then $\ln(3)+\ln(10^6)\approx1.0986 + 13.8156=14.9142$.
And $t=\frac{\ln(3\times10^6)}{\ln(2)}=\frac{\ln(3)+6\ln(10)}{\ln(2)}\approx\frac{1.0986 + 13.8156}{0.6931}\approx\frac{14.9142}{0.6931}\approx21.52$.

Step3: Solve by trial - and - error

We can also solve by calculating powers of 2:
$2^{20}=1048576$, $2^{21}=2097152$, $2^{22}=4194304$.
We can interpolate. Let $y = 2^t$. We have two points $(21,2097152)$ and $(22,4194304)$. Using the linear interpolation formula $t - t_1=\frac{y - y_1}{y_2 - y_1}(t_2 - t_1)$. Here, $y = 3\times10^6$, $y_1 = 2097152$, $y_2 = 4194304$, $t_1 = 21$, $t_2 = 22$.
$\frac{y - y_1}{y_2 - y_1}=\frac{3\times10^6 - 2097152}{4194304 - 2097152}=\frac{902848}{2097152}\approx0.43$. So $t\approx21 + 0.43=21.43$.

Answer:

The number of time - steps $t$ for the bacteria count to reach 3 million is approximately $t\approx21.52$ (using logarithms) or $t\approx21.43$ (using trial - and - error and interpolation).