Example 1
Completion requirements
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Graeme works in a laboratory investigating E. coli. He starts a new colony of bacteria with one cell in a petri dish. Every \(20\) minutes the population doubles. Graeme needs a population of at least \(15\thinspace 000\) bacteria to do his next experiment. How long will he need to wait?
Step 1: Write down what is given.
\(\begin{align}
t_1 &= 1 \\
t_n &= 15\thinspace 000 \\
r &= 2 \\
n &= ? \\
\end{align}\)
Step 2: Choose a method to solve the problem.
Use the general term to solve for \(n \).
Step 3: Solve the problem.
\(\begin{align}
t_n &= t_1 r^{n - 1} \\
15\thinspace 000 &= 1\left( 2 \right)^{n - 1} \\
15\thinspace 000 &= 2^{n - 1} \\
\end{align}\)
At this point, you do not have the tools to solve this question directly, but you can use a guess and check method to help estimate the value of \(n\). Using a base of \(2\), try various exponents, and compare the value to \(15\thinspace 000\). The goal is to find an exponent that will give a value of at least \(15\thinspace 000\).
\(\begin{align}
2^{10} &= 1\thinspace 024 &...{\rm too\thinspace low} \\
2^{12} &= 4\thinspace 096 &...{\rm {too\thinspace low}} \\
2^{13} &= 8\thinspace 192 &...{\rm too\thinspace low,\thinspace but\thinspace close...} \\
2^{14} &= 16\thinspace 384 &...{\rm this\thinspace gives\thinspace at\thinspace least\thinspace } 15\thinspace 000 \\
\end{align}\)
Use \(2^{14} \), where you left off above.
\(2^{14} = 2^{n - 1} \)
Because the bases are the same, the exponents must be the same as well, or
\(14 = n – 1 \)
\(15 = n \)
Graeme must wait \(15\) sets of \(20\) minutes before the population will be over \(15\thinspace 000\), which is \(300\) minutes, or \(5\) hours.
\(\begin{align}
t_1 &= 1 \\
t_n &= 15\thinspace 000 \\
r &= 2 \\
n &= ? \\
\end{align}\)
Step 2: Choose a method to solve the problem.
Use the general term to solve for \(n \).
Step 3: Solve the problem.
\(\begin{align}
t_n &= t_1 r^{n - 1} \\
15\thinspace 000 &= 1\left( 2 \right)^{n - 1} \\
15\thinspace 000 &= 2^{n - 1} \\
\end{align}\)
At this point, you do not have the tools to solve this question directly, but you can use a guess and check method to help estimate the value of \(n\). Using a base of \(2\), try various exponents, and compare the value to \(15\thinspace 000\). The goal is to find an exponent that will give a value of at least \(15\thinspace 000\).
\(\begin{align}
2^{10} &= 1\thinspace 024 &...{\rm too\thinspace low} \\
2^{12} &= 4\thinspace 096 &...{\rm {too\thinspace low}} \\
2^{13} &= 8\thinspace 192 &...{\rm too\thinspace low,\thinspace but\thinspace close...} \\
2^{14} &= 16\thinspace 384 &...{\rm this\thinspace gives\thinspace at\thinspace least\thinspace } 15\thinspace 000 \\
\end{align}\)
Use \(2^{14} \), where you left off above.
\(2^{14} = 2^{n - 1} \)
Because the bases are the same, the exponents must be the same as well, or
\(14 = n – 1 \)
\(15 = n \)
Graeme must wait \(15\) sets of \(20\) minutes before the population will be over \(15\thinspace 000\), which is \(300\) minutes, or \(5\) hours.