// Numbas version: exam_results_page_options {"name": "Exponentials and logarithms", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": "0", "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", ""], "variable_overrides": [[], []], "questions": [{"name": "Exponential and logarithms: Introduction - Radioactive Decay", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Evi Papadaki", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18113/"}], "tags": [], "metadata": {"description": "

Knowing the half-life of Carbon-14 and the initial mass of Carbon-14 when a tree was cut (a) write an expression that describes the relationship between the remaining mass and time, (b) calculate the remaining mass after $t$ years, and (c) given the remaining mass calculate how many years ago the tree was cut down.

", "licence": "None specified"}, "statement": "

Carbon-14 has a half-life of $5730$ years.
When a tree was cut down, it contained $\\var{m}$ $\\mathrm{gr}$ of carbon-14.

", "advice": "

a) We need to find an expression that connects the number of years and the mass of Carbon-14.

We know that Carbon-14 has a half-life of $5730$ years.We also know that the initial mass of Carbon-14 when the tree was cut down was $\\var{m}$ grams.

Let's say it takes $t$ years for the mass Carbon-14 to reach $m$ grams. Since the mass of Carbon-14 halves every $5730$ years, this means that in $t$ years the mass would have halved $\\frac{t}{5730}$ times.

So, our expression would be

\\[ \\begin{split} m&=\\var{m}\\times \\left( \\frac{1}{2} \\right) ^\\frac{t}{5730} \\qquad \\text{or} \\qquad m&=\\var{m}\\times0.5^\\frac{t}{5730}\\end{split}\\]

b) To find how much Carbon-14 the tree contained after $\\var{t1}$ years, we need to use the expression found in part (a) (either of the suggested forms would work) and set $t=\\var{t1}$:

\\[ \\begin{split} m&=\\var{m}\\times 0.5 ^\\frac{\\var{t1}}{5730} \\\\ &=\\var{m}\\times \\simplify{0.5 ^{t1/5730}}\\\\ &=\\var{m}\\times \\simplify{{0.5 ^adv1}} \\\\ &=\\simplify{{m}*{0.5 ^adv1}} \\\\&=\\var{ansb} ~~\\text{[2 d.p.]} \\end{split}\\]

Therefore, the tree contained $\\var{ansb}$ grams of Carbon-14 after $\\var{t1}$ years.

c) To find how many years ago the tree was cut if it now contains $\\var{m1}$ grams of Carbon-14, we need to use the expression found in part (a) (either of the suggested forms would work), set $m=\\var{m1}$ and then rearrange to calculate:

\\[ \\begin{split} \\var{m1}&=\\var{m}\\times 0.5 ^\\frac{t}{5730} \\\\ \\frac{\\var{m1}}{\\var{m}}&=0.5 ^\\frac{t}{5730} \\\\ \\simplify{{m1/m}}&=0.5 ^\\frac{t}{5730} \\end{split}\\]

Taking $\\log_{0.5}( )$ of both sides:

\\[ \\begin{split} \\log_{0.5}\\left(\\simplify{{m1/m}}\\right)&=\\frac{t}{5730} \\\\ 5730 \\times\\log_{0.5}\\left(\\simplify{{m1/m}}\\right)&=t \\\\ \\var{adv2}&=t \\end{split}\\]

Therefore, the tree was cut approximately $\\var{ansc}$ years ago.

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Write down an expression for the mass of carbon-14, $m$ $\\mathrm{gr}$, in the tree $t$ years after it was cut down.

$m=$[[0]] $\\mathrm{gr}$

How much carbon-14 did the tree contain after $\\var{t1}$ years?

[[0]]

Give your answer in grams to 2 decimal places if needed.

The tree now contains $\\var{m1}$ grams of carbon-14. How many years ago was it cut down?

[[0]]

Give your answer to the nearest year, if needed.

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Knowing the doubling time of a population and the population on day $t$, calculate the initial population and the number of days required for the population to reach a threshold.

", "licence": "None specified"}, "statement": "

The doubling time of a population of flies is $\\var{d}$ days.

$\\var{days[j]}$ days after the start of an experiment, there were $\\var{d1}$ flies in the population.

", "advice": "

a) We know that $\\var{ndays[j]}$ days after the start of the experience there are $\\var{d1}$ flies. We also know that the doubling time is ${\\var{d}}$ days.

In those $\\var{ndays[j]}$ days, the populations would have doubled $\\frac{\\var{ndays[j]}}{\\var{d}}=\\simplify{{ndays[j]}/{d}}$ times. Which means that the initial population, let's call it $n_0$, would have been multiplied by $2^{\\var{int}}$ to result in $\\var{d1}$ flies. So,

\\[ \\begin{split} \\simplify[!all]{n_0*2^{ndays[j]/d}}&=\\var{d1} \\\\ \\simplify{n_0*2^{ndays[j]/d}}&=\\var{d1} \\end{split} \\]

If we solve the equation we created, we can find the initial population of flies. Therefore,

\\[ \\begin{split} \\simplify{n_0*2^{ndays[j]/d}}&= \\var{d1} \\\\ n_0 &=\\frac{\\var{d1}}{\\var{2^int}} \\end{split} \\]

Which means that there are $n_0=\\frac{\\var{d1}}{\\var{2^int}}=\\var{n0}~ $ flies at the start of the experiment.

b) To calculate the number of days for the population to reach $\\var{n1}$ flies, we need an expression that connects the number of days and the size of the population.

Let's say it takes $t$ days for the fly population to reach $\\var{n1}$. Since the populatioin has doubling time of $\\var{d}$ days, this means that in $t$ days the population would have doubled $ \\frac{t}{\\var{d}}$ times. In other words, the initial population will have been multiplied by $2^{\\frac{t}{\\var{d}}}$.

So, we can now substitute this expression into our growth equation:

\\[ \\begin{split} \\var{n0}\\times 2^{\\frac{t}{\\var{d}}} &= \\var{n1} \\end{split} \\]

Now, if we solve the equaction for t we will find the number of days.

\\[ \\begin{split} \\var{n0}\\times 2^{\\frac{t}{\\var{d}}} &= \\var{n1} \\\\ 2^{\\frac{t}{\\var{d}}} &= \\frac{ \\var{n1}}{\\var{n0} } \\\\ 2^{\\frac{t}{\\var{d}}} &= \\var{n1divn0} \\end{split}\\]

Taking $log_2()$ of both sides:

\\[ \\begin{split} \\\\ {\\frac{t}{\\var{d}}} &= \\log_2(\\var{n1divn0}) \\\\ t &= \\var{d}\\times\\log_2(\\var{n1divn0}) \\\\ t &= \\var{adviceB} \\end{split}\\].

Therefore, it takes approximately $\\var{ansb}$ days for the population to reach $\\var{n1}$ flies.

Note: Alternatively, you might already know the formula the formula

\\[n=n_0 \\times 2^\\frac{t}{T} \\]

Where, $n$ represents the size of the population (e.g., number of flies) at time $t$, $n_0$ the initial population and $T$ is the length of each time step.

If so, you can use it to find the same results.

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Calculate the number of flies in the population at the start of the experiment.

There were [[0]]flies.

How many days after the start of the experiment did the population reach $\\var{n1}$ flies?

The population reached $\\var{n1}$ flies in [[0]] days.

If needed round your answer to the nearest integer.

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