Answer key · Unit 12

Exponents and logarithms.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Growth by multiplying

An amount multiplied by the same number at every step: why ten doublings give more than a thousand, how to write it as a power, and why such growth always overtakes a steady addition in the end.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Ten doublings
Practice · Predict, then look
Question
The amount starts at $1$ ⁠ and doubles at every step. How much will it be after ten steps?

Correct answer
- Correct $1024$ ⁠
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×fixed
Steps
3fixed
Ordinary scale / Log scale
Ordinary scalehidden
Choices
- Correct $1024$ ⁠
- Wrong $20$ ⁠
  Misconception: Reads doubling as adding the same amount each step: ten doublings of 1 give 20, or each doubling adds 2.
  Feedback from the catalogue
  Doubling is multiplying by $2$ ⁠, not adding $2$ ⁠. Each step adds as much as there already is, so the addition itself grows: $1, 2, 4, 8, 16, \ldots$ ⁠ After ten steps there is already $2^{10} = 1024$ ⁠. Adding the same number at every step is a linear function, a straight line, and that is a different story.
- Wrong $100$ ⁠
  Feedback
  $100$ ⁠ is $10 \times 10$ ⁠, ten squared. Here you multiply $2$ ⁠ by itself ten times.
- Wrong $512$ ⁠
  Feedback
  $512 = 2^9$ ⁠, nine steps. Count again: after one step there are $2$ ⁠, and after two $4$ ⁠.
Once the prediction is locked in, the Scene changes
Steps
10fixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$2^{10} = 1024$ ⁠. After three steps there were only $8$ ⁠, but each step doubles everything gathered so far, and the last column is already over a thousand. The amber slab on top, what the tenth step added, is half the whole column.

Worked solution
1. Start at $1$ ⁠, and multiply by $2$ ⁠ at every step.
2. After each step: $2, 4, 8, 16, 32, 64, 128, 256, 512, 1024$ ⁠.
3. For short: $2^{10} = 1024$ ⁠.
If you chose $20$ ⁠, you added $2$ ⁠ at every step instead of multiplying by $2$ ⁠. That is linear growth, $2 \times 10$ ⁠, not doubling.
Step 2 of 5
Open this step
Powers
Practice · Calculate
Question
Growth per step is locked at $\times 3$ ⁠ and Steps at $3$ ⁠, so the last column is $3^3$ ⁠. How much will there be after one more step? That is, what is $3^4$ ⁠?

Correct answer
81 ± 0.01
Setup, choices and feedback
The Scene at this step
Growth per step
3 ×fixed
Steps
3fixed
Ordinary scale / Log scale
Ordinary scalehidden
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Growth: the last column, g^n/nodes/board/outputs/growthValue
Wrong answers it recognises
- 12 ± 0.01
  Misconception: Reads a power as a product of the base and the exponent: 3⁴ = 3 × 4 = 12, or ×1.1 ten times = ×11.
  Feedback from the catalogue
  The exponent counts how many times you multiply the base by itself: $3^4 = 3 \times 3 \times 3 \times 3 = 81$ ⁠. $3 \times 4 = 12$ ⁠ is four threes added together, growth by a steady addition. In the same way, ten steps of $\times 1.1$ ⁠ make $1.1^{10}$ ⁠, not $1.1 \times 10$ ⁠.
Feedback for a right answer
$3 \times 3 = 9$ ⁠, $9 \times 3 = 27$ ⁠, $27 \times 3 = 81$ ⁠. The last column shows $3^3 = 27$ ⁠, and one more step multiplies it by $3$ ⁠: $3^4 = 3^3 \times 3$ ⁠.

Hint
$3^4 = 3 \times 3 \times 3 \times 3$ ⁠. Multiply one at a time.

Worked solution
1. $3^4$ ⁠ means four multiplications by $3$ ⁠, starting from $1$ ⁠.
2. $1 \times 3 = 3$ ⁠, $3 \times 3 = 9$ ⁠, $9 \times 3 = 27$ ⁠, $27 \times 3 = 81$ ⁠.
3. So $3^4 = 81$ ⁠.
If you got $12$ ⁠, you multiplied the base by the exponent, $3 \times 4$ ⁠. The exponent counts how many times you multiply by $3$ ⁠; it is not a number you multiply by.
Step 3 of 5
Open this step
Each step adds more
Practice · Calculate
Question
The amount doubles at every step, and there are $6$ ⁠ steps. By how much did the sixth step increase it? In other words, how big is the amber slab in the last column?

Correct answer
32 ± 0.01
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×fixed
Steps
6fixed
Ordinary scale / Log scale
Ordinary scalehidden
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Growth: what the last step added, g^n − g^(n−1) (the newest slab)/nodes/board/outputs/growthAdded
Wrong answers it recognises
- 2 ± 0.01
  Misconception: Reads doubling as adding the same amount each step: ten doublings of 1 give 20, or each doubling adds 2.
  Feedback from the catalogue
  Doubling is multiplying by $2$ ⁠, not adding $2$ ⁠. Each step adds as much as there already is, so the addition itself grows: $1, 2, 4, 8, 16, \ldots$ ⁠ After ten steps there is already $2^{10} = 1024$ ⁠. Adding the same number at every step is a linear function, a straight line, and that is a different story.
Feedback for a right answer
$2^6 - 2^5 = 64 - 32 = 32$ ⁠: the sixth step added exactly as much as everything before it. When doubling, the addition itself doubles at every step: $1, 2, 4, 8, 16, 32$ ⁠.

Hint
Compare the value above the last column with the value above the column before it.

Worked solution
1. Before the sixth step, after five steps, the amount is $2^5 = 32$ ⁠.
2. After the sixth step it is $2^6 = 64$ ⁠.
3. The sixth step added the difference: $64 - 32 = 32$ ⁠.
If you got $2$ ⁠, you thought each doubling adds $2$ ⁠. It adds as much as there already is, so the addition grows from step to step.
Step 4 of 5
Open this step
Passing a thousand
Practice · Calculate
Question
The amount starts at $1$ ⁠ and is multiplied by $3$ ⁠ at every step. At which step does it first pass $1000$ ⁠?

Correct answer
7 ± 0.01
Setup, choices and feedback
The Scene at this step
Growth per step
3 ×fixed
Steps
1
Ordinary scale / Log scale
Ordinary scalehidden
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 333 ± 1.5
  Misconception: Reads doubling as adding the same amount each step: ten doublings of 1 give 20, or each doubling adds 2.
  Feedback
  Multiplying by $3$ ⁠ is not adding $3$ ⁠. Each step multiplies everything there already is, so the addition grows from step to step: $1, 3, 9, 27, 81, \ldots$ ⁠ Adding $3$ ⁠ at every step would reach a thousand only after more than $300$ ⁠ steps.
Feedback for a right answer
$3^6 = 729$ ⁠ is not there yet, and $3^7 = 2187$ ⁠ is already over. Doubling took ten steps; with a bigger factor, a thousand comes sooner.

Feedback for a wrong answer
Not yet. At that step the value above the last column must pass $1000$ ⁠, and at the step before it must not.

Hint
Add one step at a time with Steps, and read the value above the last column.

Worked solution
1. Start at $1$ ⁠ and multiply by $3$ ⁠: $3, 9, 27, 81, 243, 729$ ⁠ after six steps.
2. $729$ ⁠ is still less than $1000$ ⁠.
3. At the seventh step: $3^7 = 729 \times 3 = 2187 > 1000$ ⁠.
4. So it is the seventh step. Set Steps to $7$ ⁠ and check.
If you got $333$ ⁠ or $334$ ⁠, you added $3$ ⁠ at every step: $1 + 3 \times 333 = 1000$ ⁠. When multiplying by $3$ ⁠, each step adds twice what there already is, so a thousand comes after only seven steps.
Step 5 of 5
Open this step
Steady addition or doubling
Practice · Choose
Question
Plan A starts from $0$ ⁠ and adds $100$ ⁠ at every step. Plan B starts from $1$ ⁠ and doubles at every step. Which is bigger after $12$ ⁠ steps?

Correct answer
- Correct Plan B, the doubling
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×fixed
Steps
6
Ordinary scale / Log scale
Ordinary scalehidden
Choices
Shown in a random order.
- Correct Plan B, the doubling
- Wrong Plan A, because it adds $100$ ⁠ at every step and the doubling starts from only $1$ ⁠
  Misconception: Thinks a large steady addition always stays ahead of doubling that starts small.
  Feedback from the catalogue
  A steady addition grows by the same amount at every step, a straight line. Doubling adds everything there already is at each step, so its addition keeps growing without end. At first it looks tiny, but once its addition passes the steady one it never falls behind again: $2^{12} = 4096$ ⁠ against $12 \times 100 = 1200$ ⁠.
- Wrong They are equal
  Feedback
  After $12$ ⁠ steps plan A is at $1200$ ⁠. Check what the board shows at step $12$ ⁠.
Feedback for a right answer
Plan A: $12 \times 100 = 1200$ ⁠. Plan B: $2^{12} = 4096$ ⁠. The doubling overtakes at the tenth step, $1024$ ⁠ against $1000$ ⁠, and from there the gap only grows: at step $20$ ⁠ it holds over a million, and plan A only $2000$ ⁠.

Worked solution
1. Plan A is a straight line, $y = 100x$ ⁠: after $12$ ⁠ steps, $100 \times 12 = 1200$ ⁠.
2. Plan B is $y = 2^x$ ⁠: after $12$ ⁠ steps, $2^{12} = 4096$ ⁠. Set Steps to $12$ ⁠ and check.
3. $4096 > 1200$ ⁠, so plan B is bigger.
If you chose plan A, you compared only the start. Up to the ninth step it does lead, but doubling adds more and more at every step, while the steady addition stays at $100$ ⁠.

Lesson 2 of 4 · About 20 minutes

Growth rates

Growth by so many per cent at every step is multiplying by a fixed factor: how to turn a percentage into a factor, why ten steps of ten per cent are more than a hundred per cent, how long it takes to double, and what happens when an amount shrinks by a percentage.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
From per cent to factor
Practice · Calculate
Question
Savings grow by $5\%$ ⁠ each year. What are they multiplied by each year?

Correct answer
1.05 ± 0.001
Setup, choices and feedback
The Scene at this step
Growth per step
1.5 ×
Steps
5
Ordinary scale / Log scale
Ordinary scalehidden
Accepted within
± 0.001
A fixed value in the question
Wrong answers it recognises
- 0.05 ± 0.001
  Misconception: Uses the percentage itself as the multiplier: 5% growth is ×0.05 (or ×5), and a 20% loss is ×0.2.
  Feedback from the catalogue
  The factor is what is left after the change, out of the whole amount. A $5\%$ ⁠ rise leaves $100\% + 5\% = 105\%$ ⁠, so you multiply by $1.05$ ⁠. A $20\%$ ⁠ fall leaves $80\%$ ⁠, so you multiply by $0.8$ ⁠. Multiplying by $0.05$ ⁠ would leave only a twentieth of the amount.
- 5 ± 0.001
  Misconception: Uses the percentage itself as the multiplier: 5% growth is ×0.05 (or ×5), and a 20% loss is ×0.2.
  Feedback from the catalogue
  The factor is what is left after the change, out of the whole amount. A $5\%$ ⁠ rise leaves $100\% + 5\% = 105\%$ ⁠, so you multiply by $1.05$ ⁠. A $20\%$ ⁠ fall leaves $80\%$ ⁠, so you multiply by $0.8$ ⁠. Multiplying by $0.05$ ⁠ would leave only a twentieth of the amount.
Feedback for a right answer
$100\% + 5\% = 105\% = 1.05$ ⁠. Each year you multiply by $1.05$ ⁠, and after $n$ ⁠ years the savings have grown $1.05^n$ ⁠-fold. Set Growth per step to $\times 1.05$ ⁠ and see how gentle the curve is.

Hint
After a year there is $100\%$ ⁠ of the savings plus $5\%$ ⁠. What is that as a decimal?

Worked solution
1. After a year all the savings are still there, $100\%$ ⁠, and the interest is added, $5\%$ ⁠.
2. Together that is $105\%$ ⁠ of the savings.
3. $105\% = \frac{105}{100} = 1.05$ ⁠, so the factor is $1.05$ ⁠.
If you wrote $0.05$ ⁠, that is only the part that is added. Multiplying by $0.05$ ⁠ would leave a twentieth of the savings.
Step 2 of 5
Open this step
Ten steps of ten per cent
Practice · Predict, then look
Question
At every step the amount grows by $10\%$ ⁠. How many times bigger will it be after ten steps?

Correct answer
- Correct About $2.6$ ⁠ times, a rise of about $160\%$ ⁠
Setup, choices and feedback
The Scene at this step
Growth per step
1.1 ×fixed
Steps
1fixed
Ordinary scale / Log scale
Ordinary scalehidden
Choices
- Correct About $2.6$ ⁠ times, a rise of about $160\%$ ⁠
- Wrong Exactly twice: ten times $10\%$ ⁠ is $100\%$ ⁠
  Misconception: Adds percentage rises over several steps: 10% for 10 steps is a 100% rise, so the amount doubles in 10 steps.
  Feedback from the catalogue
  Each percentage is worked out from the amount at that moment, which has already grown. So each step adds a little more than the one before: $1.1^{10} \approx 2.59$ ⁠, a rise of about $159\%$ ⁠, not $100\%$ ⁠. The amount has already doubled after $8$ ⁠ steps, because $1.1^{8} \approx 2.14$ ⁠.
- Wrong $11$ ⁠ times, because $1.1 \times 10 = 11$ ⁠
  Misconception: Reads a power as a product of the base and the exponent: 3⁴ = 3 × 4 = 12, or ×1.1 ten times = ×11.
  Feedback from the catalogue
  The exponent counts how many times you multiply the base by itself: $3^4 = 3 \times 3 \times 3 \times 3 = 81$ ⁠. $3 \times 4 = 12$ ⁠ is four threes added together, growth by a steady addition. In the same way, ten steps of $\times 1.1$ ⁠ make $1.1^{10}$ ⁠, not $1.1 \times 10$ ⁠.
Once the prediction is locked in, the Scene changes
Steps
10fixed
The Scene then shows: 2.594 · Growth: the last column, g^n/nodes/board/outputs/growthValue
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$1.1^{10} \approx 2.59$ ⁠. Each step adds $10\%$ ⁠ of an amount that has already grown, so the additions keep growing: the first is $0.1$ ⁠, and the last is already almost $0.24$ ⁠.

Worked solution
1. A $10\%$ ⁠ rise is multiplying by $1.1$ ⁠.
2. Ten steps are ten multiplications: $1.1^{10}$ ⁠.
3. In stages: $1.1^2 = 1.21$ ⁠, $1.1^4 = 1.21^2 \approx 1.46$ ⁠, $1.1^8 \approx 1.46^2 \approx 2.14$ ⁠, and $1.1^{10} = 1.1^8 \cdot 1.1^2 \approx 2.14 \cdot 1.21 \approx 2.59$ ⁠.
If you chose twice, you added the percentages. But each $10\%$ ⁠ is worked out from a larger amount than the last, so the result is more.
Step 3 of 5
Open this step
Doubling in five steps
Practice · Reach the target
Question
Steps is locked at $5$ ⁠. Set Growth per step so that the amount doubles in five steps: the value above the last column must be between $1.9$ ⁠ and $2.1$ ⁠. Press Check.

Goal
- 2 ± 0.1
  At the start of the step: 1.276
Setup, choices and feedback
The Scene at this step
Growth per step
1.05 ×
Steps
5fixed
Ordinary scale / Log scale
Ordinary scalehidden
Read from the Scene at
Growth: the last column, g^n/nodes/board/outputs/growthValue
One setting that reaches it
Growth per step
1.15 ×
Feedback for a right answer
$1.15^5 \approx 2.01$ ⁠: growth of $15\%$ ⁠ per step doubles in five steps. Not $20\%$ ⁠, even though $5 \times 20\% = 100\%$ ⁠: the percentages pile up on an amount that is growing.

Feedback for a wrong answer
Not yet. If the value is above $2.1$ ⁠, lower the factor; if it is below $1.9$ ⁠, raise it.

Hint
$\times 1.2$ ⁠ five times already gives almost $2.5$ ⁠. Try a smaller factor.

Worked solution
1. You need a factor $g$ ⁠ with $g^5 \approx 2$ ⁠.
2. Try $\times 1.2$ ⁠: $1.2^5 \approx 2.49$ ⁠, too big.
3. Try $\times 1.1$ ⁠: $1.1^5 \approx 1.61$ ⁠, too small.
4. In between, $\times 1.15$ ⁠: $1.15^5 \approx 2.01$ ⁠. Set Growth per step to $\times 1.15$ ⁠.
Splitting $100\%$ ⁠ into five steps gives $20\%$ ⁠, but that is already $2.5$ ⁠-fold.
Step 4 of 5
Open this step
Doubling time
Practice · Calculate
Question
Growth per step is locked at $\times 1.1$ ⁠. After how many whole steps has the amount first doubled, that is, passed $2$ ⁠?

Correct answer
8 ± 0.01
Setup, choices and feedback
The Scene at this step
Growth per step
1.1 ×fixed
Steps
1
Ordinary scale / Log scale
Ordinary scalehidden
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 10 ± 0.01
  Misconception: Adds percentage rises over several steps: 10% for 10 steps is a 100% rise, so the amount doubles in 10 steps.
  Feedback from the catalogue
  Each percentage is worked out from the amount at that moment, which has already grown. So each step adds a little more than the one before: $1.1^{10} \approx 2.59$ ⁠, a rise of about $159\%$ ⁠, not $100\%$ ⁠. The amount has already doubled after $8$ ⁠ steps, because $1.1^{8} \approx 2.14$ ⁠.
Feedback for a right answer
$1.1^7 \approx 1.95$ ⁠ is not there yet, and $1.1^8 \approx 2.14$ ⁠ is already over. A rule of thumb: the doubling time is about $70$ ⁠ divided by the percentage, here $\frac{70}{10} = 7$ ⁠ and a bit.

Hint
Move Steps and read the value above the last column.

Worked solution
1. After $n$ ⁠ steps the amount is $1.1^n$ ⁠, and you need the first $n$ ⁠ with $1.1^n > 2$ ⁠.
2. On the board: $1.1^7 \approx 1.95$ ⁠, still below $2$ ⁠.
3. $1.1^8 \approx 2.14$ ⁠, already above.
4. So $8$ ⁠ steps.
If you got $10$ ⁠, you added ten lots of $10\%$ ⁠. But the percentages pile up on a growing amount, so the doubling comes sooner.
Step 5 of 5
Open this step
When the amount shrinks
Practice · Choose
Question
A car loses $20\%$ ⁠ of its value every year. What part of the original value is left after $5$ ⁠ years?

Correct answer
- Correct About a third, because $0.8^5 \approx 0.33$ ⁠
Setup, choices and feedback
The Scene at this step
Growth per step
1.25 ×
Steps
5
Ordinary scale / Log scale
Ordinary scalehidden
Choices
Shown in a random order.
- Correct About a third, because $0.8^5 \approx 0.33$ ⁠
- Wrong Nothing: five times $20\%$ ⁠ is $100\%$ ⁠
  Misconception: Thinks a 20% loss each year takes the same amount every year, so everything is gone after 5 years.
  Feedback from the catalogue
  A percentage fall is worked out from the amount that is left, so it shrinks from year to year: each year you multiply by $0.8$ ⁠. After $5$ ⁠ years $0.8^5 \approx 0.33$ ⁠ is left, about a third. An amount multiplied at each step by a number between $0$ ⁠ and $1$ ⁠ keeps shrinking, but it never reaches zero.
- Wrong Almost nothing, because $0.2^5 = 0.00032$ ⁠
  Misconception: Uses the percentage itself as the multiplier: 5% growth is ×0.05 (or ×5), and a 20% loss is ×0.2.
  Feedback from the catalogue
  The factor is what is left after the change, out of the whole amount. A $5\%$ ⁠ rise leaves $100\% + 5\% = 105\%$ ⁠, so you multiply by $1.05$ ⁠. A $20\%$ ⁠ fall leaves $80\%$ ⁠, so you multiply by $0.8$ ⁠. Multiplying by $0.05$ ⁠ would leave only a twentieth of the amount.
Feedback for a right answer
Each year $80\%$ ⁠ of what there was is left, so you multiply by $0.8$ ⁠: $0.8^5 \approx 0.33$ ⁠. The board shows the same thing turned round: $1.25^5 \approx 3.05$ ⁠, and a third is about $\frac{1}{3.05}$ ⁠.

Worked solution
1. A $20\%$ ⁠ fall leaves $80\%$ ⁠, so each year you multiply by $0.8$ ⁠.
2. Five years: $0.8^5$ ⁠.
3. $0.8^2 = 0.64$ ⁠, $0.8^4 = 0.64^2 \approx 0.41$ ⁠, $0.8^5 \approx 0.41 \cdot 0.8 \approx 0.33$ ⁠.
4. About a third of the value is left.
If you chose “nothing”, you took $20\%$ ⁠ of the original value off each year. But each year loses $20\%$ ⁠ of what is left, an amount that keeps shrinking.

Lesson 3 of 4 · About 25 minutes

Logarithms

A logarithm answers the question "how many times do you multiply?": logarithms to base two and base ten, the log scale on which exponential growth is a straight line, and why the logarithm of a product is a sum.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
How many times?
Practice · Reach the target
Question
What is $\log_2 64$ ⁠? Set Steps so that the value above the last column is $64$ ⁠, then press Check.

Goal
- 64 ± 0.5
  At the start of the step: 2
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×fixed
Steps
1
Ordinary scale / Log scale
Ordinary scalefixed
Read from the Scene at
Growth: the last column, g^n/nodes/board/outputs/growthValue
One setting that reaches it
Steps
6
Feedback for a right answer
Six doublings: $2^6 = 64$ ⁠, so $\log_2 64 = 6$ ⁠. The logarithm is the number of steps, and the power is the value above the column.

Feedback for a wrong answer
Not yet. The value above the last column must be exactly $64$ ⁠.

Hint
$2, 4, 8, 16, \ldots$ ⁠ Count the doublings up to $64$ ⁠.

Worked solution
1. $\log_2 64$ ⁠ asks: how many times do you multiply $2$ ⁠ by itself to reach $64$ ⁠?
2. $2, 4, 8, 16, 32, 64$ ⁠: six doublings.
3. Set Steps to $6$ ⁠. $2^6 = 64$ ⁠, so $\log_2 64 = 6$ ⁠.
Step 2 of 6
Open this step
The log scale
Practice · Predict, then look
Question
When height is measured on a log scale, how will the doubling curve look?

Correct answer
- Correct A straight line
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×fixed
Steps
10fixed
Ordinary scale / Log scale
Ordinary scalefixed
Choices
- Correct A straight line
- Wrong An even steeper curve
  Feedback
  The log scale squeezes large values together rather than stretching them: the distance from $100$ ⁠ to $1000$ ⁠ equals the distance from $1$ ⁠ to $10$ ⁠.
- Wrong A flat line, because logarithms are small
  Feedback
  The height still rises at every step: each doubling climbs another $\log_{10} 2 \approx 0.3$ ⁠ of a power of ten.
Once the prediction is locked in, the Scene changes
Ordinary scale / Log scale
Log scalefixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Every step multiplies by the same factor, and on a log scale multiplying by the same factor is always the same height. The same height at every step is a straight line, as with a linear function, and its slope is $\log_{10} 2 \approx 0.301$ ⁠ per step.

Worked solution
1. On a log scale, the height of a value is its logarithm.
2. After $n$ ⁠ doublings the value is $2^n$ ⁠, and its logarithm is $\log_{10} 2^n = n \cdot \log_{10} 2$ ⁠.
3. That is a linear function of $n$ ⁠, with slope $\log_{10} 2 \approx 0.301$ ⁠: a straight line.
That is why exponential growth is drawn on a log scale: a straight line is easy to read, and its slope gives the rate of growth.
Step 3 of 6
Open this step
Powers of ten
Practice · Calculate
Question
What is $\log_{10} 1000$ ⁠?

Correct answer
3 ± 0.01
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×fixed
Steps
10fixed
Ordinary scale / Log scale
Log scale
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 100 ± 0.01
  Misconception: Reads log₁₀ 1000 as 1000 ÷ 10, or the number of doublings to a million as a million ÷ 2.
  Feedback from the catalogue
  A logarithm asks how many times you multiply: $\log_{10} 1000$ ⁠ is the number of times you multiply $10$ ⁠ by itself to reach $1000$ ⁠. $10 \times 10 \times 10 = 1000$ ⁠, so the answer is $3$ ⁠. Dividing by $10$ ⁠ gives $100$ ⁠, a different number altogether.
Feedback for a right answer
$10^3 = 1000$ ⁠, so $\log_{10} 1000 = 3$ ⁠. On the log scale $1000$ ⁠ sits three equal gaps above $1$ ⁠: one to $10$ ⁠, one to $100$ ⁠ and one to $1000$ ⁠.

Hint
How many times do you multiply $10$ ⁠ by itself to reach $1000$ ⁠? On the board, count the gaps from $1$ ⁠ up to $1000$ ⁠.

Worked solution
1. $\log_{10} 1000$ ⁠ asks: how many times do you multiply $10$ ⁠ by itself to reach $1000$ ⁠?
2. $10 \times 10 \times 10 = 1000$ ⁠: three times.
3. So $\log_{10} 1000 = 3$ ⁠. You can also count the zeros: $1000$ ⁠ has three.
If you got $100$ ⁠, you divided $1000$ ⁠ by $10$ ⁠. The logarithm is the number of multiplications, not the result of a division.
Step 4 of 6
Open this step
Every step climbs the same height
Practice · Calculate
Question
Growth per step is locked at $\times 3$ ⁠ and Steps at $4$ ⁠, so the last column is $3^4 = 81$ ⁠. The board shows how many powers of ten each step climbs. What is $\log_{10} 81$ ⁠? Round to two decimal places.

Correct answer
1.908 ± 0.01
Setup, choices and feedback
The Scene at this step
Growth per step
3 ×fixed
Steps
4fixed
Ordinary scale / Log scale
Log scalefixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Growth: log₁₀ of the last column, n · log₁₀ g/nodes/board/outputs/growthLog
Feedback for a right answer
$\log_{10} 81 = \log_{10} 3^4 = 4 \cdot \log_{10} 3 \approx 4 \cdot 0.477 \approx 1.91$ ⁠. Between $1$ ⁠ and $2$ ⁠, as it should be for a two-digit number.

Hint
Each step climbs $\log_{10} 3 \approx 0.477$ ⁠. How far do four steps climb?

Worked solution
1. $81 = 3^4$ ⁠: four steps of $\times 3$ ⁠.
2. On the log scale each step climbs $\log_{10} 3 \approx 0.477$ ⁠, as the board shows.
3. Four steps: $4 \times 0.477 = 1.908$ ⁠.
4. So $\log_{10} 81 \approx 1.91$ ⁠.
A sanity check: $10^1 = 10 < 81 < 100 = 10^2$ ⁠, so the logarithm is between $1$ ⁠ and $2$ ⁠. Answering $0.48$ ⁠ reads only one step.
Step 5 of 6
Open this step
The logarithm of a product
Practice · Choose
Question
$\log_{10} 2 \approx 0.301$ ⁠ and $\log_{10} 3 \approx 0.477$ ⁠. What is $\log_{10} 6$ ⁠?

Correct answer
- Correct $0.778$ ⁠
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×
Steps
1
Ordinary scale / Log scale
Log scale
Choices
Shown in a random order.
- Correct $0.778$ ⁠
- Wrong $0.699$ ⁠, because $\log 2 + \log 3 = \log 5$ ⁠
  Misconception: Thinks log(a + b) = log a + log b, so log 2 + log 3 = log 5.
  Feedback from the catalogue
  A logarithm turns multiplication into addition, not addition into addition: $\log(a \cdot b) = \log a + \log b$ ⁠. So $\log 2 + \log 3 = \log 6$ ⁠, because $2 \times 3 = 6$ ⁠. On the log scale each slab is a multiplication, and slabs stacked on each other add up to a height that is the sum of the logarithms.
- Wrong $0.144$ ⁠, because $0.301 \times 0.477 \approx 0.144$ ⁠
  Misconception: Thinks log(a · b) = log a · log b.
  Feedback from the catalogue
  The logarithm of a product is the sum of the logarithms: $\log(2 \cdot 3) = \log 2 + \log 3$ ⁠. The product of two logarithms smaller than $1$ ⁠ is smaller than either of them, but $6$ ⁠ is greater than both $2$ ⁠ and $3$ ⁠, so its logarithm is greater than both of theirs too.
Feedback for a right answer
$6 = 2 \cdot 3$ ⁠, so $\log_{10} 6 = \log_{10} 2 + \log_{10} 3 \approx 0.301 + 0.477 = 0.778$ ⁠. On the board: a slab of $\times 2$ ⁠ with a slab of $\times 3$ ⁠ on top reach the height of $\times 6$ ⁠ together.

Worked solution
1. Write $6$ ⁠ as a product: $6 = 2 \cdot 3$ ⁠.
2. The product rule: $\log(a \cdot b) = \log a + \log b$ ⁠.
3. $\log_{10} 6 = \log_{10} 2 + \log_{10} 3 \approx 0.301 + 0.477 = 0.778$ ⁠.
If you chose $0.699$ ⁠, you took the sum of the logarithms to be the logarithm of the sum, $2 + 3 = 5$ ⁠. A logarithm turns multiplication into addition, so the sum belongs to the product, $6$ ⁠.
Step 6 of 6
Open this step
Solving with a logarithm
Practice · Calculate
Question
How many doublings, starting from $1$ ⁠, does it take to pass a million? That is, what is the first whole number $n$ ⁠ with $2^n > 10^6$ ⁠?

Correct answer
20 ± 0.01
Setup, choices and feedback
The Scene at this step
Growth per step
2 ×fixed
Steps
10
Ordinary scale / Log scale
Log scale
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 500000 ± 0.01
  Misconception: Reads log₁₀ 1000 as 1000 ÷ 10, or the number of doublings to a million as a million ÷ 2.
  Feedback from the catalogue
  A logarithm asks how many times you multiply: $\log_{10} 1000$ ⁠ is the number of times you multiply $10$ ⁠ by itself to reach $1000$ ⁠. $10 \times 10 \times 10 = 1000$ ⁠, so the answer is $3$ ⁠. Dividing by $10$ ⁠ gives $100$ ⁠, a different number altogether.
Feedback for a right answer
$\frac{6}{0.301} \approx 19.9$ ⁠, so $20$ ⁠ doublings: $2^{19} = 524288$ ⁠ is not there yet, and $2^{20} = 1048576$ ⁠ is already over. In other words, $\log_2 10^6 \approx 19.9$ ⁠.

Hint
A million is $6$ ⁠ powers of ten, and each doubling climbs about $0.301$ ⁠ of them. How many doublings climb $6$ ⁠?

Worked solution
1. You need $n$ ⁠ with $2^n > 10^6$ ⁠. Take $\log_{10}$ ⁠ of both sides: $n \cdot \log_{10} 2 > 6$ ⁠.
2. $\log_{10} 2 \approx 0.301$ ⁠, so $n > \frac{6}{0.301} \approx 19.9$ ⁠.
3. The first whole number is $20$ ⁠. A check: $2^{10} = 1024$ ⁠, so $2^{20} = 1024^2 = 1048576$ ⁠, just over a million.
If you got $500000$ ⁠, you divided a million by $2$ ⁠. But the question is how many times you double, and that is a question for a logarithm.

Lesson 4 of 4 · About 20 minutes

Decay and half-life

Radioactive decay is exponential decay: in every half-life, half of what was there is left. How much is left after several half-lives, how long it takes to reach a given fraction, and why a small sample does not give exact numbers.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A decaying sample
Practice · Predict, then look
Question
The sample’s half-life is 5 s. What part of it will be left after two half-lives, that is, after 10 s?

Correct answer
- Correct A quarter
Setup, choices and feedback
The Scene at this step
Half-life
5 s
Nuclei in the sample
400
Show the decay curve
Off
Choices
- Correct A quarter
- Wrong Nothing: half decays in the first and the other half in the second
  Misconception: Thinks two half-lives make a whole life: half goes in the first, the other half in the second, and nothing is left.
  Feedback from the catalogue
  In each half-life, half of what is left goes, not half of the starting amount. After the first, $\frac{1}{2}$ ⁠ is left; after the second, half of that, $\frac{1}{4}$ ⁠; after the third, $\frac{1}{8}$ ⁠. That is multiplying by $\frac{1}{2}$ ⁠ each time, so the amount keeps shrinking and never reaches zero.
- Wrong Half, because the half-life does not change
  Feedback
  The half-life does stay the same, so in each one half of what is left goes again. After the second, less than half is left.
Once the prediction is locked in, the Scene changes
Show the decay curve
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
In the first half-life half decays, and $\frac{1}{2}$ ⁠ is left. In the second, half of what is left decays, and $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ ⁠ is left. On the graph, the orange line passes near $\frac{1}{4}$ ⁠ above $2$ ⁠.

Worked solution
1. In each half-life, what is left is multiplied by $\frac{1}{2}$ ⁠.
2. After the first: $\frac{1}{2}$ ⁠.
3. After the second: $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ ⁠.
If you chose “nothing”, in the second you took off half of the starting amount. But the second halves what is left, which is a quarter of the start.
Step 2 of 5
Open this step
Halving again
Practice · Calculate
Question
What part of the sample is left after three half-lives? Write it as a decimal.

Correct answer
0.125 ± 0.002
Setup, choices and feedback
The Scene at this step
Half-life
5 s
Nuclei in the sample
400
Show the decay curve
Off
Accepted within
± 0.002
A fixed value in the question
Wrong answers it recognises
- 0 ± 0.001
  Misconception: Thinks two half-lives make a whole life: half goes in the first, the other half in the second, and nothing is left.
  Feedback from the catalogue
  In each half-life, half of what is left goes, not half of the starting amount. After the first, $\frac{1}{2}$ ⁠ is left; after the second, half of that, $\frac{1}{4}$ ⁠; after the third, $\frac{1}{8}$ ⁠. That is multiplying by $\frac{1}{2}$ ⁠ each time, so the amount keeps shrinking and never reaches zero.
- 0.1667 ± 0.005
  Misconception: Thinks the fraction left after n half-lives is 1/(2n), dividing by 2 times n instead of by 2 to the power n.
  Feedback from the catalogue
  Each half-life multiplies what is left by $\frac{1}{2}$ ⁠. After $3$ ⁠ of them: $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2^3} = \frac{1}{8}$ ⁠. $\frac{1}{2 \cdot 3} = \frac{1}{6}$ ⁠ mixes up multiplying by two with a power of two.
Feedback for a right answer
$\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125$ ⁠. Choose On under Show the decay curve: above $3$ ⁠ the white curve passes exactly through $\frac{1}{8}$ ⁠.

Hint
Half, then half of that, then half again.

Worked solution
1. Each half-life multiplies by $\frac{1}{2}$ ⁠.
2. Three of them: $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ ⁠.
3. $\frac{1}{8} = 1 \div 8 = 0.125$ ⁠.
If you got $0.167$ ⁠, you divided by $2 \cdot 3 = 6$ ⁠. But three halvings divide by $2^3 = 8$ ⁠.
Step 3 of 5
Open this step
How long?
Practice · Calculate
Question
Half-life is locked at 4 s. How long does it take until an eighth of the sample is left?

Correct answer
12 s ± 0.1 s
Setup, choices and feedback
The Scene at this step
Half-life
4 sfixed
Nuclei in the sample
400
Show the decay curve
On
Accepted within
± 0.1 s
Computed from the Scene at this step’s setting
Half life/nodes/sample/props/halfLife
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 32 s ± 0.1 s
  Misconception: Thinks it takes 8 half-lives to get down to an eighth.
  Feedback from the catalogue
  An eighth is half of half of half: $\frac{1}{8} = \left(\frac{1}{2}\right)^3$ ⁠. So you reach it after three half-lives, not eight. After eight only $\frac{1}{256}$ ⁠ is left. The number of half-lives is the exponent, that is, the logarithm to base $2$ ⁠.
Feedback for a right answer
$\frac{1}{8} = \left(\frac{1}{2}\right)^3$ ⁠, so it takes three half-lives: $3 \times 4 = 12$ ⁠ seconds.

Hint
How many half-lives does it take to reach $\frac{1}{8}$ ⁠?

Worked solution
1. An eighth is half of half of half: $\frac{1}{8} = \left(\frac{1}{2}\right)^3$ ⁠, that is three half-lives.
2. Each lasts 4 s.
3. $3 \times 4 = 12$ ⁠, that is 12 s.
If you got 32 s, you thought an eighth takes eight half-lives. After eight only $\frac{1}{256}$ ⁠ is left.
Step 4 of 5
Open this step
A handful of nuclei
Practice · Choose
Question
The sample holds $20$ ⁠ nuclei, and the half-life is 5 s. How many nuclei will be left after 5 s?

Correct answer
- Correct About $10$ ⁠, but it could be $7$ ⁠ or $13$ ⁠
Setup, choices and feedback
The Scene at this step
Half-life
5 s
Nuclei in the sample
20fixed
Show the decay curve
Off
Choices
Shown in a random order.
- Correct About $10$ ⁠, but it could be $7$ ⁠ or $13$ ⁠
- Wrong Exactly $10$ ⁠: that is what half-life means
  Misconception: Thinks exactly half of any sample decays in each half-life, even a handful of nuclei.
  Feedback from the catalogue
  Each nucleus decays at a random moment, whatever its neighbours do. The half-life says each nucleus has an even chance of decaying in that time, like a coin. Twenty coin tosses do not always give ten heads: you might get $7$ ⁠ or $13$ ⁠. Only in a large sample does the randomness average out, so that very nearly half decay.
- Wrong There is no saying at all
  Feedback
  There is no knowing when a particular nucleus will decay, but there is plenty to say about the whole sample: on average half decay, give or take.
Feedback for a right answer
Each nucleus is like a coin: an even chance of decaying within the half-life. Twenty tosses usually give about ten, but not always exactly. In a sample of $400$ ⁠ the spread is much smaller compared with the sample, which is why the orange line there stays almost on the curve.

Worked solution
1. The half-life says each nucleus has an even chance of decaying in that time.
2. With $20$ ⁠ nuclei, the average is $20 \cdot \frac{1}{2} = 10$ ⁠.
3. But each nucleus draws its own lot, so the result spreads around $10$ ⁠: $7$ ⁠ or $13$ ⁠ are not rare.
If you chose “exactly $10$ ⁠”, you treated the half-life as a timetable. It holds on average, and only in a large sample does it look exact.
Step 5 of 5
Open this step
Under one per cent
Practice · Calculate
Question
After how many whole half-lives is less than $1\%$ ⁠ of the sample first left?

Correct answer
7 ± 0.01
Setup, choices and feedback
The Scene at this step
Half-life
2 s
Nuclei in the sample
400
Show the decay curve
On
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 2 ± 0.01
  Misconception: Thinks two half-lives make a whole life: half goes in the first, the other half in the second, and nothing is left.
  Feedback from the catalogue
  In each half-life, half of what is left goes, not half of the starting amount. After the first, $\frac{1}{2}$ ⁠ is left; after the second, half of that, $\frac{1}{4}$ ⁠; after the third, $\frac{1}{8}$ ⁠. That is multiplying by $\frac{1}{2}$ ⁠ each time, so the amount keeps shrinking and never reaches zero.
- 50 ± 0.01
  Misconception: Thinks the fraction left after n half-lives is 1/(2n), dividing by 2 times n instead of by 2 to the power n.
  Feedback from the catalogue
  Each half-life multiplies what is left by $\frac{1}{2}$ ⁠. After $3$ ⁠ of them: $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2^3} = \frac{1}{8}$ ⁠. $\frac{1}{2 \cdot 3} = \frac{1}{6}$ ⁠ mixes up multiplying by two with a power of two.
Feedback for a right answer
After six, $\frac{1}{64} \approx 1.6\%$ ⁠ is left, and after seven $\frac{1}{128} \approx 0.8\%$ ⁠. In the words of logarithms: $\log_2 100 \approx 6.6$ ⁠, rounded up to $7$ ⁠.

Hint
Look for the first power of $2$ ⁠ greater than $100$ ⁠: $2, 4, 8, 16, \ldots$ ⁠

Worked solution
1. After $n$ ⁠ half-lives $\frac{1}{2^n}$ ⁠ is left, and you need $\frac{1}{2^n} < \frac{1}{100}$ ⁠, that is $2^n > 100$ ⁠.
2. $2^6 = 64$ ⁠, still less than $100$ ⁠.
3. $2^7 = 128 > 100$ ⁠.
4. So $7$ ⁠ half-lives.
If you got $2$ ⁠, you thought two half-lives remove everything. After two a quarter, $25\%$ ⁠, is still left.

Across the unit · 13 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Growth by multiplying

Ten doublings

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Powers

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Each step adds more

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Passing a thousand

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Steady addition or doubling

Question

Correct answer

The Scene at this step

Choices

Worked solution

Growth rates

From per cent to factor

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Ten steps of ten per cent

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Doubling in five steps

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Doubling time

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

When the amount shrinks

Question

Correct answer

The Scene at this step

Choices

Worked solution

Logarithms

How many times?

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The log scale

Question