Estimated
world human population
from 10,000 BCE to 2017 CE. For most of human history the count crept upward; then, in the last few centuries, it began doubling faster and faster. If a population doubles each year, today’s question is the exponent in 1 · 2x = 1,000,000. Image: El T / Wikimedia Commons —
File:Population_curve.svg, public domain.
If a population doubles every year
and currently stands at one million people,
when did it have just one?
Standards: HSF-LE.A.4 (preview). This lesson motivates logarithms — students should leave wanting a better way than guessing. Do not introduce log notation today; that is L9. Prep checklist: scientific calculators required; graphing tech recommended; print of trapezoid pattern image (or project from slide 4). The title image is a world-population curve — use it to anchor the Big Question: doubling is the engine that turns one ancestor into a million, and the unknown is how many doublings.
Warm-Up · Do Now5 min
Do Now
Complete on your own. Then confirm your answers with a partner.
Solve \(x + 8 = 20\).
Solve \(7x = 28\).
Solve \(x^{2} = 49\).
Solve \(x^{3} = 27\).
Solve \(x - 5 = 11\).
Solve \(\sqrt{x} = 9\).
🖨️ Print Worksheet — Do Now
Listen for the three solving moves students use — (1) a familiar math fact (“I just know \(7 \cdot 4 = 28\)”), (2) working backward from the result, and (3) applying an inverse operation (subtract, divide, square root, cube root). These three moves are the scaffold for today's lesson.
Foreshadow: when the unknown is an exponent instead of the base, only working backward still works cleanly — we don't yet have an inverse operation in our toolkit. That discomfort is the entry point to Activity 1.
Watch for: Problem 3 — some students give only \(x = 7\); the complete answer is \(x = \pm 7\). Don't push on this now; surface it in the synthesis if it comes up.
Warm-Up · Synthesis~2 min
Three ways to undo an operation
Each equation asked you to undo an operation to find the unknown. There are three moves we keep coming back to.
Inverse Operation
Apply the opposite operation to both sides.
e.g.
\[\begin{aligned}
x + 8 &= 20 \\
x + 8 - 8 &= 20 - 8 \\
x &= 12
\end{aligned}\]
Root (undo a power)
Take the matching root of both sides.
e.g.
\[\begin{aligned}
x^{3} &= 27 \\
\sqrt[3]{x^{3}} &= \sqrt[3]{27} \\
x &= 3
\end{aligned}\]
Rewrite as same base
Match the bases, then match the exponents.
e.g.
\[\begin{aligned}
2^{x} &= 32 \\
2^{x} &= 2^{5} \\
x &= 5
\end{aligned}\]
“In which of today's equations did ‘undo by inverse operation’ fail to be clean?”None of today's equations — every one had a clean inverse: subtract, divide, or take a root. But if the unknown were the exponent (like \(2^{x} = 32\) when the target isn't a nice power of 2), the inverse operation isn't in our toolkit yet. Rewriting as the same base only works when the target happens to be a power of the base. That gap is what today's lesson opens up.
click to advance discussion ▶
Try Saying
To undo ___, I can use ___ because ___.
Try Saying
If the unknown were the exponent instead of the base, this move would/wouldn't still work because ___.
Hand-off (~2 min). Use this slide to name the three moves explicitly — students have done all three, but they may not have a name for each. Anchor the warm-up's \(x^{3} = 64\) row under Root, and the \(2^{x} = 32\) row under Rewrite as same base.
End by saying: “Today we'll meet equations where the unknown is the exponent. None of our three moves quite finishes the job — we'll have to estimate.” Hand off to Activity 1.
The fourth move — the logarithm — is named in the next lesson. Do not pre-teach it here.
Activity 1 · Launch3 min
A Tessellated Trapezoid
A single trapezoid is split into 4 smaller, similar trapezoids. At the next step, each of those splits into 4 again. The pattern continues without end. How many smallest trapezoids do we see at each step?
Each step splits every trapezoid into 4 similar trapezoids.
How many small trapezoids appear at Step 1? At Step 2? What pattern do you notice in the count?
Try Saying
At step ___, there are ___ trapezoids, because each previous trapezoid splits into ___.
Launch (~3 min) · Three Reads. Read 1 — What is happening? A single tile keeps subdividing into similar pieces. Read 2 — What is being counted? The number of smallest trapezoids visible at each step (not the total across steps). Read 3 — What pattern do you notice in the count? Anchor on the move: each step multiplies the previous count by 4. Do not reveal the formula \(f(n) = 4^n\) yet — let students conjecture it during work time. If a student writes “\(4 \cdot n\)” on the board, hold that as a productive misconception to be tested against Step 2 (\(4 \cdot 2 = 8 \neq 16\)).
Activity 1 · Work Time10 min
A Tessellated Trapezoid
Each step splits every trapezoid into 4 similar trapezoids. The count follows the pattern 1, 4, 16, 64, … — multiplying by 4 at every step.
If n is the step number, how many smallest trapezoids appear when \(n = 4\)? When \(n = 6\)?
Write an equation relating the number of trapezoids T to the step n.
At a certain step, there are 65,536 trapezoids. What is n?
Explain to a partner how you found n.
10 min · Activity 1 Work Time. Leave this slide on screen during work time.
Monitor for the five strategies in the activity narrative: (1) repeat-multiply — just keep multiplying by 4 and counting; (2) calculator guess-and-check on \(4^n\); (3) skip-ahead from a known power (e.g., start at \(4^4 = 256\) and continue); (4) table or spreadsheet of \(n\) vs. \(4^n\); (5) graph \(y = 4^x\) and read off where \(y = 262{,}144\).
Select 2–3 strategies to share at the synthesis (next slide). Aim for one slow-but-reliable (repeat-multiply) and one faster-with-tools (calculator/table/graph).
Watch for students who multiply the base by the step number (\(4 \cdot 7\) instead of \(4^7\)) — this is the linear vs. exponential confusion that motivates the whole lesson. Anchor on the splitting picture: every step, every existing trapezoid splits into 4. So at step \(n\) you have multiplied by 4 a total of \(n\) times.
Amplify · 4 inputs
Order: 4 separate Text→Response→Note sequences, named answer1a, answer1b, answer2, answer3. Problem 4 (explain to a partner) is verbal — no input.
Format examples: Example: 256 · Example: 4096 · Example: T = 4^n · Example: 8
answer2 is a Text Response (free-form equation). The other three are Math Responses graded on numericValue.
Problem 1a: Number of trapezoids when n = 4
content:
when answer1a.submitted and answer1a.numericValue > 255.5 and answer1a.numericValue < 256.5
"Correct! Step 4 means four splittings of the original. 4 · 4 · 4 · 4 = 256 trapezoids."
when answer1a.submitted and answer1a.numericValue > 15.5 and answer1a.numericValue < 16.5
"Almost! 16 is the count at step 2, not step 4. Step 4 means four splittings of the original: 4 · 4 · 4 · 4."
when answer1a.submitted and answer1a.numericValue > 63.5 and answer1a.numericValue < 64.5
"You're close — 64 is step 3 (4^3). Step 4 multiplies once more by 4."
when answer1a.submitted and answer1a.numericValue > 1023.5 and answer1a.numericValue < 1024.5
"Hmm... 1024 is 4^5. Step 4 is one multiplication earlier."
when answer1a.submitted and not isBlank(answer1a.latex)
"Take another look — at each step every trapezoid splits into 4. After 4 splits, the count is 4 · 4 · 4 · 4. What is that?"
otherwise ""
Problem 1b: Number of trapezoids when n = 6
content:
when answer1b.submitted and answer1b.numericValue > 4095.5 and answer1b.numericValue < 4096.5
"Correct! Starting from 4^4 = 256, multiply by 4 two more times: 256 · 4 · 4 = 4,096."
when answer1b.submitted and answer1b.numericValue > 16383.5 and answer1b.numericValue < 16384.5
"Almost — 16,384 is 4^7, one step too far. Back off by one multiplication by 4."
when answer1b.submitted and answer1b.numericValue > 1023.5 and answer1b.numericValue < 1024.5
"Almost — 1,024 is 4^5, one step short. Multiply once more by 4."
when answer1b.submitted and answer1b.numericValue > 23.5 and answer1b.numericValue < 24.5
"Not quite — 24 looks like 4 · 6. The count is multiplicative, not a product. At step 6 you have multiplied by 4 six times."
when answer1b.submitted and answer1b.numericValue > 35.5 and answer1b.numericValue < 36.5
"I see what happened — 36 is 6^2. The base is 4, and the exponent is the step number. Try 4^6."
when answer1b.submitted and not isBlank(answer1b.latex)
"Good start — extend your pattern. Starting from 4^4 = 256, multiply by 4 two more times. What do you get?"
otherwise ""
Problem 2: Equation relating T to step n
content:
when answer2.submitted and answer2.content matches "T\\s*=\\s*4\\s*\\^\\s*\\{?\\s*n\\s*\\}?"
"Correct! Each step multiplies the previous count by 4, so after n steps the count is 4^n. Equation: T = 4^n."
when answer2.submitted and answer2.content matches "4\\s*\\^\\s*\\{?\\s*n\\s*\\}?"
"You're close — that captures the right rule. The full equation names both sides: T = 4^n, where T is the count and n is the step."
when answer2.submitted and isBlank(answer2.content)
"Hmm... take another look at the pattern: step 0 gives 1, step 1 gives 4, step 2 gives 16. At step n you have multiplied by 4 a total of n times. Write that as an exponent."
when answer2.submitted
"Good start — at step 0 there is 1 trapezoid. At step 1 there are 4. At step n you have multiplied by 4 a total of n times. Write that as an exponent: T = 4^?."
otherwise ""
Problem 3: Value of n when T = 65,536
content:
when answer3.submitted and answer3.numericValue > 7.99 and answer3.numericValue < 8.01
"Correct! 4^8 = 65,536. Multiplying 4 by itself 8 times — or starting from 4^4 = 256 and multiplying by 4 four more times — lands exactly on the target."
when answer3.submitted and answer3.numericValue > 6.99 and answer3.numericValue < 7.01
"Almost — 4^7 = 16,384, which is too small. Try one more multiplication by 4."
when answer3.submitted and answer3.numericValue > 8.99 and answer3.numericValue < 9.01
"Hmm — 4^9 = 262,144, four times too big. One less multiplication."
when answer3.submitted and answer3.numericValue > 65535.5 and answer3.numericValue < 65536.5
"I see what happened — 65,536 is the *value* T. The question asks for the *exponent* n that gives T = 65,536."
when answer3.submitted and not isBlank(answer3.latex)
"You're close — multiply 256 (which is 4^4) by 4 repeatedly. Count how many times it takes to reach 65,536."
otherwise ""
Activity 1 · Synthesis5 min
Tessellated Trapezoid — Strategies Share
There is more than one way to find an unknown exponent. Some strategies are slow but reliable; others are fast but need extra tools. Surface the strategies students used — then ask which one fits the question.
“What is the same across the strategies students used?”Every strategy is looking for the value of \(n\) that makes \(4^{n} = 65{,}536\). They all rely on knowing the base (4) and the target (65,536). The only thing that changes is how we hunt for the exponent — repeated multiplication, calculator guess-and-check, a table, or a graph.
“Which strategy would you use if the target were small, like \(4^{n} = 64\)?”Repeat-multiply or recall: \(4^{1} = 4,\ 4^{2} = 16,\ 4^{3} = 64\), so \(n = 3\). Fast, exact, no calculator needed. When the answer is a small integer, knowing the powers of the base is the quickest move.
“Which strategy would you use if the target were huge, like \(4^{n} = 268{,}435{,}456\)?”Calculator with a table of powers, or a graph of \(y = 4^{x}\). Repeat-multiply still works, but counting 14 multiplications by hand is slow and error-prone. (Answer: \(n = 14\).) We want a faster move.
click to advance discussion ▶
Try Saying
For this target, I would use ___ because ___, and not ___ because ___.
Run as Compare-and-Connect (MLR7). Select 2–3 students you watched during work time who used different strategies — ideally one repeat-multiplier, one calculator-guess-and-checker, and one table or graph user. Have each name their strategy in one sentence, then steer the class to the comparison questions above.
End by writing \(4^{n} = 65{,}536\) on the board and asking: “Is there a name for the operation that gives us \(n\) directly — the way \(\sqrt{\;}\) gave us the base in \(x^{2} = 49\)?” Pause. Don't answer. Tell students the name is coming in the next lesson; for now, the move is called finding the unknown exponent. This is the hand-off into the MASL slide.
Math As A Second LanguageVisual / MASL3 min
Equations with an Unknown Exponent
Math · We Say · Meaning
★ Math (given)
\(b^{x} = y\)
▲ We Say
“b to the x equals y”
or
“b raised to the unknown power x equals y”
● Meaning
An exponential equation where the exponent is the unknown. To solve, ask: what power must we raise the base b to in order to land on y? Example: \(2^{x} = 32\) means \(x = 5\).
Try Saying
“b to the x equals y — the unknown is the ___, and to solve I need to ask what power of ___ lands on ___.”
Read the We Say card aloud chorally, then have one student read the Meaning card. Connect back to the trapezoid problem: in \(4^{n} = 262{,}144\), the base b is 4, the target y is 262,144, and the unknown x is the step number n — which we found to be 9. Tell students this notation (general form \(b^{x} = y\)) will be everywhere this week, and that we still don't have a clean inverse operation for it. The next slide (How To) lists the strategies we have so far; the next lesson introduces the missing operation.
Reference · How To
2 min
How To: Estimate an Unknown Exponent
Step
What to do
Example: \(b^{\,x} = y\)
1
Identify the base b and the target value y. The unknown is the exponent x.
From \(4^{\,n} = 262{,}144\): base \(b = 4\), target \(y = 262{,}144\). Unknown: the exponent n.
2
Try integer exponents first — multiply the base by itself, counting as you go.
If no integer works, sandwich the answer between two integers, then estimate decimals.
\(3^{\,t} = 10\). We know \(3^{2} = 9\) and \(3^{3} = 27\), so t is between 2 and 3, closer to 2. Test \(3^{2.1} \approx 10.05\).
★
Or graph \(y = b^{\,x}\) and read the x-value where the curve crosses y.
Graph \(y = 3^{\,x}\) and find the x-value at \(y = 10\). Reads about \(x \approx 2.10\).
🧑🏫 Leave this slide on screen during Activity 2 Work Time. Click rows one at a time — action first, then the worked example. Step 2 covers the integer-clean case (Activity 2 Part A and the Missing Values table). Step 3 covers the decimal-approximation case (Part B questions 4 and 5). The ★ row flags graphing as an alternative entry point — useful for students who reason visually. Don’t pre-teach the strategies; let students try, then point back here when they get stuck.
Activity 2 · Launch2 min
A Tripling Bacterial Colony
Enterobacter colony in a petri dish — a bacterial population can triple in a single hour.
A petri dish starts with 100 thousand bacteria. The colony triples every hour. If P is the population in thousands and t is time in hours, then:
\(P(t) = 100 \cdot 3^{\,t}\)
How would you find the population at \(t = 2\)? Why is finding the population at \(t = 1.5\) a different kind of question?
Try Saying
To find the population after ___ hours, I would ___. This is the same/different kind of question as ___ because ___.
Quick launch (~2 min). The activity has two halves: (a) given the exponent, find the population — straightforward substitution into \(100 \cdot 3^t\); (b) given the population, find the exponent — the new move that no familiar inverse cleanly undoes. Don't telegraph that split yet — let it emerge from the work time on the next slide. Pair students before sending them in.
The discussion prompt is doing real work: \(t = 2\) is a clean substitution (\(P = 900\)); \(t = 1.5\) looks similar but requires a calculator and the same exponential rules students saw with half-lives in L7 (fractional exponent on the base 3). If a student says “just take the average of \(P(1)\) and \(P(2)\),” flag that as a linear guess — we'll revisit in synthesis.
Activity 2 · Work Time13 min
A Tripling Bacterial Colony
A colony of Enterobacter starts at 100 thousand cells. It triples every hour. After \(t\) hours, the population (in thousands) is
\( \;P(t) \;=\; 100 \cdot 3^{\,t} \). \)
On your own — then compare with a partner
Part A — Find the population
After 4 hours.
After 90 minutes (\(t = 1.5\)).
After \(\tfrac{1}{3}\) of an hour.
Part B — Find the time
When does the population reach 1,000 thousand?
When does the population double (reach 200 thousand)?
Part C — Missing values for \(2^{\,x}\)
Fill in the three highlighted cells.
\(x\)
0
1
2
?
8
?
\(2^{\,x}\)
1
2
4
16
?
1024
13 min · Activity 2 Work Time. Leave this slide and the slide-8 How-To both reachable — direct students to glance back when stuck on Part B.
Part A (substitute): \(P(4) = 100 \cdot 3^{4} = 8100\); \(P(1.5) = 100 \cdot 3^{1.5} \approx 519.6\); \(P(1/3) = 100 \cdot 3^{1/3} \approx 144.2\). Each one is a clean “plug in & evaluate.”
Part B (estimate): \(100 \cdot 3^{t} = 1000 \Rightarrow t \approx 2.10\); \(100 \cdot 3^{t} = 200 \Rightarrow t \approx 0.63\). These force estimation — the calculator can't solve directly. Sandwich, then refine with decimals.
Part C (table): \(2^{4} = 16\) so the missing top-row \(x\) is \(4\); \(2^{8} = 256\); \(2^{10} = 1024\) so the missing top-row \(x\) is \(10\).
Watch for students who reason that Part B Question 5 must give \(t < 1\) because the colony triples in one hour (so it doubles before then). That MP2 sandwich-reasoning is gold — flag it for the synthesis.
Watch for the linear-mind error: \(P(1.5) = 450\) by averaging 300 and 900. Surface that this looks reasonable but exponentials don't average that way.
Amplify · 8 inputs
Order: 8 separate Text→Math Response→Note sequences, named answer1, answer2, answer3, answer4, answer5, answer6a, answer6b, answer6c.
content:
when answer1.submitted and answer1.numericValue > 8099.5 and answer1.numericValue < 8100.5
"Correct! 100 · 3^4 = 100 · 81 = 8,100 thousand cells. Four triplings: 100 → 300 → 900 → 2,700 → 8,100."
when answer1.submitted and answer1.numericValue > 1199.5 and answer1.numericValue < 1200.5
"Almost — 1,200 looks like 100 · 3 · 4 (multiplied instead of exponentiated). The base 3 has to be raised to the 4th power. Try 100 · 3^4."
when answer1.submitted and answer1.numericValue > 80.5 and answer1.numericValue < 81.5
"Hmm — 81 is just 3^4. Don't forget the starting 100. Multiply by 100."
when answer1.submitted and answer1.numericValue > 11.5 and answer1.numericValue < 12.5
"I see what happened — 12 is 3 · 4. The base 3 needs to be raised to the 4th power, not multiplied by 4. Try 100 · 3^4."
when answer1.submitted and not isBlank(answer1.latex)
"You're close — the colony triples each hour for 4 hours. Compute 100 · 3 · 3 · 3 · 3. What do you get?"
otherwise ""
Problem A2: Population at \(t = 1.5\) hours (90 min).
content:
when answer2.submitted and answer2.numericValue > 519 and answer2.numericValue < 520.5
"Correct! 100 · 3^(1.5) ≈ 519.6 thousand cells. 3^(1.5) = 3 · √3 ≈ 5.196."
when answer2.submitted and answer2.numericValue > 449.5 and answer2.numericValue < 450.5
"Not quite — 450 is the midpoint between 300 (t = 1) and 900 (t = 2). That's a linear guess. Use the exponential: 100 · 3^(1.5)."
when answer2.submitted and answer2.numericValue > 149.5 and answer2.numericValue < 150.5
"Hmm — 150 looks like 100 · 1.5. The base 3 needs to be raised to the 1.5 power, not multiplied by 1.5."
when answer2.submitted and answer2.numericValue > 599.5 and answer2.numericValue < 600.5
"You're close — 600 is in the right neighborhood. Compute 100 · 3^(1.5) on the calculator. Try entering 100*3^(1.5)."
when answer2.submitted and not isBlank(answer2.latex)
"Good start — substitute t = 1.5 into 100 · 3^t. On a calculator: 100 * 3^(1.5). What do you get?"
otherwise ""
Problem A3: Population at \(t = 1/3\) hour.
content:
when answer3.submitted and answer3.numericValue > 143.5 and answer3.numericValue < 144.6
"Correct! 100 · 3^(1/3) ≈ 144.2 thousand cells. 3^(1/3) is the cube root of 3, about 1.442."
when answer3.submitted and answer3.numericValue > 33.2 and answer3.numericValue < 33.4
"Hmm — 33.3 looks like 100 / 3. We need to multiply by 3^(1/3), not divide by 3."
when answer3.submitted and answer3.numericValue > 100.2 and answer3.numericValue < 100.4
"Almost — 3^(1/3) ≈ 1.442, so 100 · 1.442 ≈ 144. Re-check your calculator entry: 100 * 3^(1/3)."
when answer3.submitted and not isBlank(answer3.latex)
"Take another look — try 100 * 3^(1/3) on the calculator. The cube root of 3 is about 1.44."
otherwise ""
Problem B4: Time when the population reaches 1,000 thousand.
content:
when answer4.submitted and answer4.numericValue > 2.05 and answer4.numericValue < 2.15
"Correct! t ≈ 2.10 hours. Check: 100 · 3^(2.10) ≈ 1,003 thousand cells."
when answer4.submitted and answer4.numericValue > 1.99 and answer4.numericValue < 2.01
"You're close — at t = 2, the population is 100 · 9 = 900, just under 1,000. The answer is slightly more than 2."
when answer4.submitted and answer4.numericValue > 2.99 and answer4.numericValue < 3.01
"Almost — at t = 3, the population is 100 · 27 = 2,700, way past 1,000. Try a t between 2 and 3, closer to 2."
when answer4.submitted and answer4.numericValue > 9.99 and answer4.numericValue < 10.01
"Hmm — 10 might be 1000 / 100. We need 3^t = 10, and t is the *exponent* that does it, not the ratio itself."
when answer4.submitted and not isBlank(answer4.latex)
"I see what happened — sandwich the answer: 100 · 3^2 = 900 and 100 · 3^3 = 2,700. So t is between 2 and 3. Try decimals: t = 2.1? t = 2.2?"
otherwise ""
Problem B5: Time when the population doubles to 200 thousand.
content:
when answer5.submitted and answer5.numericValue > 0.6 and answer5.numericValue < 0.66
"Correct! t ≈ 0.63 hours. Check: 100 · 3^(0.63) ≈ 199.9 thousand cells."
when answer5.submitted and answer5.numericValue > 0.49 and answer5.numericValue < 0.51
"Almost — at t = 0.5, the population is 100 · √3 ≈ 173, not yet 200. The doubling time is slightly more than 0.5."
when answer5.submitted and answer5.numericValue > 0.99 and answer5.numericValue < 1.01
"Hmm — at t = 1, the population has tripled (300), well past doubling. Doubling happens *before* one hour."
when answer5.submitted and answer5.numericValue > 1.99 and answer5.numericValue < 2.01
"Take another look — by t = 2 the population has gone up nine-fold, not doubled. Doubling happens in less than one hour."
when answer5.submitted and not isBlank(answer5.latex)
"You're close — we need 3^t = 2. Since 3^1 = 3 and 3^0 = 1, t is between 0 and 1, closer to 1. Try decimals like t = 0.6 or t = 0.7."
otherwise ""
Problem C6a: Missing top-row \(x\) when \(2^{\,x} = 16\).
content:
when answer6a.submitted and answer6a.numericValue = 4
"Correct! 2^4 = 16. The doubling chain: 1, 2, 4, 8, 16 — four doublings."
when answer6a.submitted and answer6a.numericValue = 8
"Hmm — 2^8 = 256, not 16. What power of 2 equals 16?"
when answer6a.submitted and answer6a.numericValue = 2
"Almost — 2^2 = 4, not 16. Try one more power up."
when answer6a.submitted and answer6a.numericValue = 16
"I see what happened — 16 is the *value*, not the exponent. We want the power: what x makes 2^x = 16?"
when answer6a.submitted and not isBlank(answer6a.latex)
"Good start — count: 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = ?. Which one equals 16?"
otherwise ""
Problem C6b: Missing bottom-row value when \(x = 8\).
content:
when answer6b.submitted and answer6b.numericValue = 256
"Correct! 2^8 = 256. Eight doublings: 1, 2, 4, 8, 16, 32, 64, 128, 256."
when answer6b.submitted and answer6b.numericValue = 16
"Almost — 16 is 2^4. You need 2^8. Multiply 16 by 2 four more times."
when answer6b.submitted and answer6b.numericValue = 64
"You're close — 64 is 2^6. Two more doublings."
when answer6b.submitted and answer6b.numericValue = 128
"Not quite — 128 is 2^7, one doubling short. One more multiplication by 2."
when answer6b.submitted and not isBlank(answer6b.latex)
"Take another look — double 1 eight times: 1, 2, 4, 8, 16, 32, 64, 128, 256. The eighth doubling gives you the answer."
otherwise ""
Problem C6c: Missing top-row \(x\) when \(2^{\,x} = 1024\).
content:
when answer6c.submitted and answer6c.numericValue = 10
"Correct! 2^10 = 1024. Two more doublings past 2^8 = 256: 256 → 512 → 1024."
when answer6c.submitted and answer6c.numericValue = 9
"Almost — 2^9 = 512, one short. One more doubling reaches 1024."
when answer6c.submitted and answer6c.numericValue = 1024
"Hmm — 1024 is the *value*, not the exponent. We want the power: what x makes 2^x = 1024?"
when answer6c.submitted and answer6c.numericValue = 11
"You're close — 2^11 = 2048, double 1024. Step back one exponent."
when answer6c.submitted and not isBlank(answer6c.latex)
"Good start — keep doubling: 2^8 = 256, 2^9 = 512, 2^10 = ?. Which exponent lands on 1024?"
otherwise ""
Activity 2 · Synthesis5 min
Homework Refresher: Which one can we solve? Which can we estimate?
“What strategies did you use to find t when the population was 1,000 thousand?”Multiply 100 by 3 repeatedly and count the steps. Or try decimal exponents between 2 and 3 on the calculator until the result lands near 1,000. Or graph \(y = 100 \cdot 3^{x}\) and read off the x-value where \(y = 1000\). All three give \(t \approx 2.10\) — none of them give it exactly.
“What is missing from our toolkit?”A clean inverse operation for \(3^{t}\). Just as \(\sqrt{\;}\) undoes \(x^{2}\) and subtraction undoes addition, we want an operation that undoes “3 to the power of.” That operation exists — it is called a logarithm — and it is the subject of our next lesson.
click to advance discussion ▶
Try Saying
When ___ is unknown, I can substitute. When ___ is unknown, I have to estimate, because ___.
End by writing the guiding question on the board: “Is there a better way to find the value of t than by approximating?” Tell students yes — that is exactly where logarithms come from, starting next lesson. Do not introduce the symbol “log” today; the goal here is only to make students feel the need for it. Listen for students who reason that the doubling time must satisfy \(t < 1\) because the population triples in 1 hour — that MP2 reasoning is gold.
Lesson Synthesis~3 min
Finding an unknown exponent
We met two problems today — trapezoid steps and bacteria hours — that had the same shape: \(b^x = y\), solve for \(x\). Here is the menu of moves and the open question we leave with.
\(b^x = y\)
Repeat-multiply
Multiply the base by itself; count to the target. Best when \(y\) is small and the answer is a clean integer.
Calculator + decimals
Try decimal exponents on the calculator until the result matches \(y\). Best for fast convergence to two decimals.
Sandwich + estimate
Find consecutive integers with \(b^n < y < b^{n+1}\); estimate decimals between them. Best for rough mental answers.
Graph and read off
Graph \(y = b^x\); read the \(x\)-value where the curve crosses the target \(y\). Best when a graphing tool is available.
“How is finding an unknown exponent like — and unlike — solving \(x + 3 = 10\) or \(x^2 = 16\)?”Like: we’re isolating an unknown by working backward from a known result. Unlike: for \(x+3=10\) we have subtraction, and for \(x^2=16\) we have \(\sqrt{\;}\). For \(b^x = y\), we don’t yet have an operation that cleanly undoes exponentiation — only the four estimation moves above. That missing operation is coming next lesson.
click to advance discussion ▶
Try Saying
To find an unknown exponent, I would ___. The drawback is ___.
I Can… (SWBAT)
Comprehend the need to undo exponentiation to solve problems in simple contexts.
Explain how to determine the value of an unknown exponent both exactly and approximately.
💻 Strategy Practice (digital)🖨️ Worksheet (2-sided B&W) Lesson Synthesis (~3 min). Walk through the four-strategy menu, naming each move and where it shines. Use the queue question to compare today’s equations with familiar ones — the point is that we have an unknown but no clean inverse operation yet. End by writing \(b^x = y\) on the board and saying: “Next lesson, we name the operation that undoes \(b^x\). It’s called a logarithm.” Don’t introduce the symbol today.
Strategy Practice: the linked digital worksheet has a worked example + 2 problems per method (8 total). Method 4 directs students to Desmos. Students write their answers on the print companion to hand in.
SWBAT belongs here, not on the title slide. Read aloud after the discussion concludes.
Cool-Down5 min
Video Viewers
On the day a video is posted, 5 people watch it. The next day the count doubles. The doubling continues each day, so the number of viewers on day d is given by
\[5 \cdot 2^{d}.\]
On your own — show your reasoning
On which day will 640 people see the video? Show your reasoning.
What strategy would you use to find the first day when more than 20,000 people will see the video?
Cool-Down · 5 min. Exit ticket. Students must work independently. Leave this slide on screen during exit ticket. The slide-8 How-To (Estimate an Unknown Exponent) is the paired reference for this cool-down — keep it visible or toggle back to it so students can glance at the strategy menu while writing.
Problem 1: Solve \(5 \cdot 2^{d} = 640\). Divide by 5: \(2^{d} = 128\). Count doublings: \(2^{7} = 128\), so d = 7. Equivalent: double 5 day by day — 5, 10, 20, 40, 80, 160, 320, 640 — and count the doublings (7 of them).
Problem 2: Accept any reasoned strategy. Common moves: (a) keep doubling until you pass 20,000; (b) divide 20,000 by 5 to get 4,000, then ask what power of 2 exceeds 4,000 (\(2^{12} = 4096\), so d = 12 is the first day); (c) graph \(y = 5 \cdot 2^{x}\) and read where it crosses 20,000. Actual first day is d = 12 (since \(5 \cdot 2^{12} = 20{,}480\)).
Watch for students who type the value 4,000 (= 20,000 / 5) or 20,000 for Problem 1, confusing the day (exponent) with the viewer count. Redirect: the day is the exponent in \(5 \cdot 2^{d}\), not the count itself.
Amplify · 2 inputs
Order per problem: Text component (problem text + Example: 7) → Math Response (Problem 1, named answer1) or Text Response (Problem 2, named answer2) → Note with CL.
Problem 1: numericValue check. Problem 2: answer2.content non-blank acknowledgement.
Problem 1: On which day will 640 people see the video? (Enter the day number.)
content:
when answer1.submitted and answer1.numericValue > 6.99 and answer1.numericValue < 7.01
"Correct! Solve 5 · 2^d = 640. Divide by 5: 2^d = 128. Count doublings: 2^7 = 128. Day 7."
when answer1.submitted and answer1.numericValue > 5.99 and answer1.numericValue < 6.01
"Almost — on day 6, 5 · 2^6 = 320 people watch. One more doubling reaches 640. Try one day later."
when answer1.submitted and answer1.numericValue > 7.99 and answer1.numericValue < 8.01
"Hmm — on day 8, 5 · 2^8 = 1,280 people watch. That overshoots 640. Try one day earlier."
when answer1.submitted and answer1.numericValue > 127.99 and answer1.numericValue < 128.01
"I see what happened — 128 is 640/5, the number of doublings worth of growth. The day is the exponent: what power of 2 equals 128?"
when answer1.submitted and answer1.numericValue > 639.99 and answer1.numericValue < 640.01
"Take another look — 640 is the viewer count, not the day. The day is the exponent d in 5 · 2^d = 640. Solve for d."
when answer1.submitted and isBlank(answer1.latex)
"Enter the day number. Double 5 day by day: 5, 10, 20, 40, 80, 160, 320, 640. Which day reaches 640? Example: 7"
when answer1.submitted
"Take another look — double 5 day by day: 5, 10, 20, 40, 80, 160, 320, 640. Count the doublings. Which day reaches 640?"
otherwise ""
Problem 2: What strategy would you use to find the first day when more than 20,000 people will see the video?
content:
when answer2.submitted and not(isBlank(answer2.content))
"Thank you — share your strategy in tomorrow's opener. Common moves: keep doubling 5 until you pass 20,000; divide 20,000 by 5 and ask what power of 2 exceeds 4,000 (2^12 = 4096, so day 12); or graph y = 5 · 2^x and read where it crosses 20,000."
when answer2.submitted and isBlank(answer2.content)
"Please describe a strategy before submitting. For example: 'I would keep doubling 5 each day until the count passes 20,000, then report that day.'"
otherwise ""
Lesson Summary
Unknown exponents — what we learned
A town's population was 1 thousand. It has doubled every decade and is currently 32 thousand. How many decades has it been? We can write the situation as an equation with an unknown exponent:
Rewind the clock: starting at 1,000, when was the population 250?Doubling backwards is halving, so a negative exponent works the same way.
Solve \(2^{d} = \tfrac{1}{4}\) — we get \(d = -2\). The population was 250 two decades before it was 1,000.
\[2^{d} = \tfrac{1}{4} \;\Longrightarrow\; d = -2\]
click to reveal ▶
Looking Ahead
Sometimes the exponent is easy to spot — \(2^{d} = 32\) means \(d = 5\). But what about \(2^{d} = 100\), or \(3^{d} = 100\)? There is no integer that works. To solve equations like these, we need a brand new operation — one that undoes \(b^{x}\). Next lesson: logarithms.
Lesson Summary slide — closure / homework reference. Read aloud or hand off as a reading. The big move today: students went from "evaluate \(b^x\)" (Section A) to "solve \(b^x = y\) for \(x\)" (today). Foreshadow tomorrow: the operation that undoes \(b^x\) will be named — the logarithm. Students will write \(\log_{2}(100)\) for the answer to \(2^{d} = 100\), just like they write \(\sqrt{2}\) for the answer to \(x^{2} = 2\).
