Algebra 2 · Unit 5 · Lesson 11

Evaluating Logarithmic Expressions

Black-and-white portrait photograph of seismologist Charles F. Richter, inventor of the Richter magnitude scale, in his Caltech office.
Seismologist Charles F. Richter, who in 1935 at Caltech defined the magnitude scale that still carries his name. Each whole-number step on the scale is a 10× jump in ground shaking — and roughly a 32× jump in released energy. A magnitude-9 quake (Tōhoku, 2011) is written as a single digit because the scale is logarithmic.
Image: Caltech Archives via Wikimedia Commons — File:Charles_Francis_Richter.jpg, public domain.

\(\log(1{,}000{,}000{,}000) = 9.\)

Why does a billion collapse to a single digit?

Standards: HSF-LE.A.4, HSF-BF.B.4.a. This (A)-lesson opens the formal study of logarithms as the inverse of base-10 exponentiation. Students will leave today able to (1) decide whether a log is positive, zero, or negative by inspection, (2) bracket a base-10 log between consecutive integers, and (3) use the log button on a calculator with correct input format.

Why this image: The Richter scale is the most visceral demonstration of why we need logarithms. Each whole-number step is a 10× jump in shaking amplitude — so a magnitude-9 quake releases roughly 32× the energy of an 8 but is written as a single digit. The Big Question reframes compression of scale as the structural job logarithms do.

(A)-lesson note: The IM source builds Activity 1 around a physical "Log War" card game. The SDP reframe preserves the reasoning targets (sign, integer bracket, closeness) in a slide-based "Compare Without Computing" flow. The card game is retained as an optional extension on the Activity 1 launch slide.
Prep Checklist
  • ☐ Scientific calculator for every student (the log button is the centerpiece of Activity 2).
  • ☐ Optional: printed Log War decks for the card-game extension in Activity 1.
  • ☐ No printed organizer required — this lesson runs entirely from the slide deck plus calculators.
  • ☐ Forward-link: tomorrow (Lesson 12) introduces the change-of-base form log(x)/log(b) — preview surfaces on the Activity 2 work-time slide.
Warm-Up 3 min

Do Now

Evaluate mentally. Be ready to share your reasoning.
  1. Find the value of \(\log_{2}(8)\).
  2. Find the value of \(\log_{5}(625)\).
  3. Find the value of \(\log(0.1)\).
  4. Find the value of \(\log(0.001)\).
Math Talk routine — reveal one problem at a time, collect strategies before moving on. Students work silently for ~30 seconds per problem, then share reasoning aloud.

The decimals (#3, #4) seed the idea that logs of numbers less than 1 are negative — the structural insight that makes Activity 1 (Compare Without Computing) possible.

Strategy to surface: watch for students who rewrite \(0.1 = 10^{-1}\) and \(0.001 = 10^{-3}\). That rewrite is the move that turns a hard mental computation into a one-step exponent read — name it explicitly in synthesis (slide 3).

No ACB on warm-up (discussion only). Answers and the sign rule are formalized on slide 3 synthesis.
Warm-Up · Synthesis 2 min

Logs Without a Calculator — Discussion

A logarithm asks: what exponent makes this true? When the number is bigger than 1, the answer is positive. When the number is between 0 and 1, the answer is negative.

If \(x > 1\)
\(\log_b x > 0\)
If \(x = 1\)
\(\log_b 1 = 0\)
If \(0 < x < 1\)
\(\log_b x < 0\)
“Why is \(\log_2 8 = 3\)?” Because \(2^{3} = 8\). The log asks for the exponent that takes the base to the argument — here, that exponent is 3.
“Why is \(\log 0.1 = -1\)?” Because \(0.1 = 10^{-1}\). Rewriting the decimal as a power of 10 makes the exponent visible — and the exponent is the log.
“How can you tell — before computing — whether \(\log_b x\) will be positive, negative, or zero?” For \(b > 1\): positive when \(x > 1\), zero when \(x = 1\), negative when \(0 < x < 1\). The sign of a log is determined entirely by where the argument sits relative to 1.
click to advance discussion ▶
Try Saying First, I ___ because ___. I noticed ___, so I ___.
End the warm-up with the sign rule explicit on the board: positive when the argument is greater than 1, zero when it equals 1, negative when it is between 0 and 1. Students will use this rule in the next ten minutes to compare log expressions without a calculator (slide 6).

The decimal items from the Do Now (\(\log 0.1\), \(\log 0.001\)) are the structural payoff — surface any student who rewrote \(0.1 = 10^{-1}\) or \(0.001 = 10^{-3}\). That move is the strategy.

If anyone says “log of 1 is zero in any base” — capture it on the board and reference it on slide 4 when the MASL row introduces \(\log_b 1 = 0\) as a callout.

No ACB on synthesis (discussion only).
MASL
2 min
Reading a Logarithm
Math · We Say · Meaning
★ Math (given)
\(\log_b x = n \iff b^n = x\)
▲ We Say
“log base b of x equals n if and only if b to the n equals x
● Meaning
A logarithm and an exponential equation express the same relationship. \(\log_2 8 = 3\) and \(2^3 = 8\) say exactly the same thing.
\(\log_b 1 = 0\)
If \(0 < x < 1\) and \(b > 1\), then \(\log_b x < 0\).
Read the We Say card aloud, slowly, pointing at each symbol as you say it. Then trace the equivalence on the Math card: “log base b of x — the question — equals n — the answer — exactly when b raised to n gives back x.” The two callouts below the row are the sign-rule anchors from the warm-up: \(\log_b 1 = 0\) (any valid base) and logs of numbers between 0 and 1 are negative. Students will lean on both in the next ten minutes on slide 5.
Activity 1 · Launch 3 min

Which Is Bigger?

Before We Start · 1–2 min
Think of a time you compared two things without measuring them exactly — figuring out which bag of groceries was heavier, which song was louder, which line at the store was moving faster. How did you know?
Today we are going to compare logarithms without using a calculator. The goal is not an exact value — it is a confident comparison.
For example
\(\log_2 30 \;\stackrel{?}{>}\; \log_2 7\)
Yes — because \(30 > 7\) and the log function grows as its input grows. We did not compute either value.
What clues could we use to compare two logs without a calculator? Where might we look first?
Try Saying I think ___ is greater than ___ because ___.
Launch (~3 min). Open with the violet funds-of-knowledge prompt — 1–2 minutes of partner or whole-class talk about real comparisons made without measurement. Invite 1–2 student responses, then transition into the priming example.

If no one volunteers: share your own. “When I am in a grocery aisle and have to pick a checkout line, I do not count people — I look at carts. A full cart with one person is slower than three people each holding two items. That is a comparison without computing.”

Bridge to the math: Read the priming example aloud and trace the reasoning — sign of the log, integer bracket, and closeness within a bracket are the three moves students will use on the next slide. Reference the warm-up sign rule (\(\log_b x < 0\) when \(0 < x < 1\)) explicitly — it is the easiest of the three moves and the entry point for problem 1.

Optional · Log War extension. If your class has printed Log War decks, you can run the IM card-game version of Activity 1 here (20 min) — give pairs the first 20 cards, have them order least-to-greatest, then play War. Re-join the deck at slide 7. Otherwise proceed with slide 6 as the slide-based version.
Activity 1 · Work Time 12 min

Compare Without Computing

For each pair, decide which expression is greater. No calculators. Defend your choice using sign, integer bracket, or closeness reasoning.
On your own, then compare with a partner
  1. Which is greater: \(\log_2 30\) or \(\log_2 \tfrac{1}{4}\)?  (Enter L or R.)
  2. Which is greater: \(\log 500\) or \(\log 5000\)?  (Enter L or R.)
  3. Which is greater: \(\log 980\) or \(\log 110\)?  (Enter L or R.)
Three Reasoning Moves
  • Sign — is the argument \(> 1\), \(= 1\), or between 0 and 1?
  • Bracket — between which two consecutive powers of the base?
  • Closeness — which end of the bracket is the argument nearer to?
Try Saying “\(\log\,\_\_\_\) is greater than \(\log\,\_\_\_\) because ___.”
12 min · Activity 1 Work Time. Leave this slide on screen during work time.

Monitor for two strategies. Sign-based (problem 1: positive always beats negative). Bracket-based (problems 2 and 3: which two integers does each log sit between?). The closeness move in problem 3 is the bridge to Activity 2's calculator-estimation work.

Watch for students who reach for a calculator. Redirect: “You don't need an exact value — just a confident comparison.”

Card-game extension. If your class is running printed Log War decks (see slide 5 teacher note), students do the deck-ordering task instead and you can skip the on-screen problems.

Surface for synthesis: a sign-based explanation for problem 1, a bracket-based explanation for problem 2, and a closeness-based explanation for problem 3 — one student each if possible.
Amplify · 3 inputs

Order: 3 separate Text→Math Response→Note sequences, named answer1, answer2, answer3. Each takes a one-letter answer (L if the left expression is greater, R if the right is).

Format examples: Example: L · Example: R · Example: L

Problem 1: log_2(30) vs. log_2(1/4) — which is greater? (Enter L or R)
content: when answer1.submitted and (answer1.latex matches "L" or answer1.latex matches "l") "Correct! log_2(30) is positive (since 30 > 1) and log_2(1/4) is negative (since 1/4 < 1). Positive always beats negative." when answer1.submitted and (answer1.latex matches "R" or answer1.latex matches "r") "Take another look — what is the *sign* of each log? Numbers between 0 and 1 give negative logs, and any positive log is greater than any negative log." when answer1.submitted and isBlank(answer1.latex) "Type L if the left expression is greater, R if the right is. (Example: L)" otherwise ""
Problem 2: log(500) vs. log(5000) — which is greater? (Enter L or R)
content: when answer2.submitted and (answer2.latex matches "R" or answer2.latex matches "r") "Yes — log(5000) sits between 3 and 4 (since 1000 < 5000 < 10000), while log(500) sits between 2 and 3 (since 100 < 500 < 1000). The right expression is greater." when answer2.submitted and (answer2.latex matches "L" or answer2.latex matches "l") "Almost — bracket each one. 500 is between 100 and 1000, so log(500) is between 2 and 3. Where does 5000 sit between consecutive powers of 10?" when answer2.submitted and isBlank(answer2.latex) "Type L or R. (Example: R)" otherwise ""
Problem 3: log(980) vs. log(110) — which is greater? (Enter L or R)
content: when answer3.submitted and (answer3.latex matches "L" or answer3.latex matches "l") "Correct! log(980) is just under 3 (since 980 is close to 1000), and log(110) is just above 2 (since 110 is close to 100). The left is greater by almost a whole unit." when answer3.submitted and (answer3.latex matches "R" or answer3.latex matches "r") "I see what happened — both 110 and 980 sit between 100 and 1000, so both logs are between 2 and 3. The question is *where* in that bracket each one falls. 110 is just above 100; 980 is just below 1000. Which is closer to 3?" when answer3.submitted and isBlank(answer3.latex) "Type L or R. (Example: L)" otherwise ""
Activity 1 · Synthesis 3 min

Three Ways to Compare — Without a Calculator

Every comparison we just made used one of three reasoning moves. Name the move, then we will see what the calculator says.

“\(\log_2 30\) vs. \(\log_2 \tfrac{1}{4}\) — which is greater, and how did you know without computing?” Sign. \(\log_2 30\) is positive (because \(30 > 1\)) and \(\log_2 \tfrac{1}{4}\) is negative (because \(0 < \tfrac{1}{4} < 1\)). Positive always beats negative — no arithmetic required.
“\(\log 500\) vs. \(\log 5000\) — which is greater, and what was the move?” Integer bracket. \(500\) sits between \(10^{2}\) and \(10^{3}\), so \(\log 500\) is between \(2\) and \(3\). \(5000\) sits between \(10^{3}\) and \(10^{4}\), so \(\log 5000\) is between \(3\) and \(4\). Different brackets → the higher bracket wins.
“\(\log 980\) vs. \(\log 110\) — same bracket. So what tipped the comparison?” Closeness. Both sit between \(2\) and \(3\). But \(980\) is right up against \(1000\), so \(\log 980\) is just under \(3\). And \(110\) is just past \(100\), so \(\log 110\) is just over \(2\). The left wins by almost a whole unit.
“Let's verify the trickiest one. About what value would the calculator give for \(\log 980\)?” Our estimate said “just under 3.” Type it in: log(980)  →  2.991. The estimate predicted the calculator — not the other way around.
click to advance discussion ▶
Try Saying I compared \(\log\,\_\_\_\) and \(\log\,\_\_\_\) using ___ (sign / bracket / closeness), because ___.
Synthesis (3 min). Name the three reasoning moves explicitly — sign, integer bracket, closeness — and write them on the board in that order of refinement. Each move handles a harder case than the last: sign separates by category, bracket separates by integer range, closeness separates within a bracket.

The fourth click is the bridge to the next slide: students saw their estimate for \(\log 980\) come out “just under 3,” and the calculator confirms log(980) ≈ 2.991. Read the input aloud as “log open-paren nine-eighty close-paren” — students will need to hear this syntax again on slide 8 when they enter it themselves. The estimate is the check on the calculator, never the other way around. If a student says “why not just always use the calculator?” — that is the question slide 9 (Activity 2 Synthesis) answers explicitly.
Activity 2 · Work Time 8 min

Estimate, Then Verify

Tyler and Priya both want to know the value of \(\log 19\). Tyler reaches for paper; Priya picks up her calculator. They both press the log button, type the number, and read the screen. The calculator entry looks like this:
log(19)  →  1.2787536…
The log button computes \(\log_{10}\) — base 10. Type the number in parentheses, just like a function. The screen returns a decimal that rounds to whatever precision you need.
Estimate first, then verify with your calculator
  1. Estimate \(\log 75\) using the bracket-and-closeness method. Then enter log(75) on your calculator and round to two decimal places.
  2. Estimate \(\log 1020\). Then enter log(1020) and round to two decimal places.
  3. Estimate \(\log 9\). Then enter log(9) and round to two decimal places.
A glimpse ahead
Your calculator's log button only computes base 10. If you ever need a log in a different base, the calculator will use the change-of-base form:
log(x)/log(b)
For example, \(\log_2 8\) becomes a single calculator entry: log(8)/log(2). We will use this in Lesson 12.
Try Saying I estimated ___ because ___. The calculator showed ___.
8 min · Activity 2 Work Time. Leave this slide on screen during work time. Slide 7's How To (four-step estimate) stays referenceable — project side-by-side if your setup allows.

Calculator-input format moment. Read the entry aloud: “log open-paren nineteen close-paren.” Students hear the syntax once, formally, with Tyler and Priya as anchors. The change-of-base aside is a preview only — do not work \(\log_2 8\) here; that is Lesson 12.

Expected values: \(\log 75 \approx 1.88\) (between 10 and 100, closer to 100); \(\log 1020 \approx 3.00\) (just past \(10^3\)); \(\log 9 \approx 0.95\) (just under 1, since 9 is just under 10).

Watch for the student who types log(9), reads 0.954, and writes 9.54. That misread — sliding the decimal — is the most common error and the Tyler/Priya rounding question on slide 9 is the place to surface it.
Amplify · 3 inputs

Order: 3 separate Text→Math Response→Note sequences, named answer1a, answer1b, answer1c. All numeric, ±0.01 tolerance.

Format examples: Example: 1.88 · Example: 3.01 · Example: 0.95

Problem 1a: Estimate, then compute log(75). Enter the calculator value rounded to 2 decimal places.
content: when answer1a.submitted and answer1a.numericValue > 1.87 and answer1a.numericValue < 1.89 "Correct — log 75 ≈ 1.88. Because 75 is between 10 and 100, the log lands between 1 and 2, closer to 2 since 75 is closer to 100." when answer1a.submitted and answer1a.numericValue > 1.84 and answer1a.numericValue < 1.91 "Very close — round to two decimals. The calculator shows 1.8750, which rounds to 1.88." when answer1a.submitted and answer1a.numericValue > 9.5 and answer1a.numericValue < 9.6 "I see what happened — the calculator reads 1.8750, not 18.75. Check the decimal point: log(75) ≈ 1.88, not 18.8." when answer1a.submitted and isBlank(answer1a.latex) "Type your calculator value rounded to two decimal places. (Example: 1.88)" otherwise "Hmm — bracket 75 between powers of 10. What two integers does log 75 sit between? Then enter log(75) and round to 2 decimals."
Problem 1b: Estimate, then compute log(1020). Enter the calculator value rounded to 2 decimal places.
content: when answer1b.submitted and answer1b.numericValue > 3.00 and answer1b.numericValue < 3.02 "Yes — log 1020 ≈ 3.01. Just barely above 3 because 1020 is just barely above 1000." when answer1b.submitted and answer1b.numericValue > 2.98 and answer1b.numericValue < 3.04 "Almost — keep one more decimal. The calculator shows 3.0086, which rounds to 3.01." when answer1b.submitted and answer1b.numericValue > 2.0 and answer1b.numericValue < 2.1 "Take another look — 1020 is just past 10³, not 10². Where does that put log 1020?" when answer1b.submitted and isBlank(answer1b.latex) "Type your calculator value. (Example: 3.01)" otherwise "You're close — 1020 sits just above 1000 = 10³. So log 1020 sits just above 3. Enter log(1020) and round."
Problem 1c: Estimate, then compute log(9). Enter the calculator value rounded to 2 decimal places.
content: when answer1c.submitted and answer1c.numericValue > 0.94 and answer1c.numericValue < 0.97 "Correct — log 9 ≈ 0.95. Just under 1 because 9 is just under 10, and log 10 = 1." when answer1c.submitted and answer1c.numericValue > 0.90 and answer1c.numericValue < 0.99 "Close — round to two decimals. The calculator shows 0.9542, which rounds to 0.95." when answer1c.submitted and answer1c.numericValue > 9.5 and answer1c.numericValue < 9.6 "I see what happened — the calculator reads 0.9542, not 9.54. Watch the decimal point: log(9) is less than 1." when answer1c.submitted and isBlank(answer1c.latex) "Type your calculator value. (Example: 0.95)" otherwise "Good start — 9 is just under 10, so log 9 is just under log 10 = 1. Enter log(9) and round to 2 decimals."
Activity 2 · Synthesis 3 min

Why Estimate Before You Type?

The calculator gives a number. The estimate tells you whether to trust it.

“Tyler types log(19) and reads 1.279. Priya types the same thing and reads 1.27875. Why might these differ?” Different rounding settings or display digits — nothing more. Both readings are correct to the precision shown. A quick bracket check confirms both: \(10^{1} = 10\) and \(10^{2} = 100\), so \(\log 19\) is between 1 and 2, much closer to 1 (since 19 is much closer to 10 than to 100). Both 1.279 and 1.27875 sit exactly where the estimate says they should.
“Without a calculator — is 2.675 a reasonable answer for \(\log_2 18\)?” No. Bracket the argument: \(2^{4} = 16\) and \(2^{5} = 32\), so \(\log_2 18\) must sit between 4 and 5. 2.675 is not even in the right whole-number range — someone likely typed \(\log 18\) (base 10) instead of \(\log_2 18\). The estimate caught the error before the calculator could hide it.
click to advance discussion ▶
Try Saying My estimate is ___. The calculator gives ___. The two agree / don’t agree because ___.
Quick discussion (~3 min). The big takeaway: the estimate is the check on the calculator, not the other way around. Calculators round, drop digits, and accept whatever base-10 keystroke you give them — even when you meant a different base.

Q1 normalizes precision differences (Tyler vs. Priya). Many students assume one of two slightly-different calculator readings must be “wrong” — surface that both are correct, and that the bracket between 1 and 2 confirms both.

Q2 is the “is this reasonable?” check that the lesson summary depends on. The 2.675 answer is the classic base-mistake artifact: it is what you get when you compute \(\log_{10} 18\) on a calculator that has no base-2 button. Without the estimate, the student would accept it. With the estimate, they catch it instantly.

End with this rule on the board: “When the calculator answer and your estimate disagree by more than a tenth or so — believe the estimate first and re-enter the calculation.”

No ACB on synthesis (discussion only).
Lesson Synthesis ~3 min

What We Can Do Now

A logarithm asks: "what power of the base gives me this number?" We have four moves — two recall facts, one estimation routine, and one calculator step. Pick the one that fits the question.

“Without typing anything: is \(\log_5 1\) positive, zero, or negative? What about \(\log_5 0.2\)?” \(\log_5 1 = 0\) (any base raised to the 0 power is 1). \(\log_5 0.2\) is negative, because 0.2 is less than 1 — you need a negative exponent on 5 to get a number below 1. (Specifically, \(5^{-1} = 0.2\), so \(\log_5 0.2 = -1\).)
click to reveal ▶
Try Saying To find a logarithm, I can ______. The exact value is hard to find when ______, so I estimate by ______.
I Can… (SWBAT)
I can recall \(\log_b 1 = 0\) and \(\log_b b = 1\) for any valid base.
I can decide whether a log is positive, zero, or negative just by looking at the number inside it.
I can estimate the value of a base-10 log by bracketing the number between consecutive powers of 10.
I can use the log button on a calculator to find the value of any base-10 log.
Lesson Synthesis (~3 min). Cold-call the question and let students reason out loud before clicking the answer. The point is that the two recall facts (\(\log_b 1 = 0\), \(\log_b b = 1\)) plus the sign check (is the input above, equal to, or below 1?) handle a surprising number of logs without any computation at all. Estimation by bracketing and the calculator's log button cover the rest.

Read the SWBAT block aloud after the discussion concludes — this is confirmation of what students just did today, not a forecast of what they will do. SWBAT belongs here, not on the title slide.
Cool-Down 5 min

Cool-Down — Calculating Logs

Express the solution to each equation as a logarithm. Then use your calculator to find the approximate value, rounded to four decimal places.
On your own — exit ticket
  1. Solve \(10^x = 32\). Express \(x\) as a logarithm, then compute.
  2. Solve \(10^x = 0.25\). Express \(x\) as a logarithm, then compute.
Cool-Down · 5 min. Independent — no partner work. Leave this slide on screen during exit ticket.

Slide 7 (How To) remains accessible via the nav if students get stuck on the rewrite step \(10^x = N \Rightarrow x = \log N\).

Diagnostic: Problem 2's sign is what to watch. Students who confidently produce a negative answer have internalized the warm-up sign rule (\(\log_b x < 0\) when \(0 < x < 1\)). A positive answer on problem 2 is the flag for reteach tomorrow.

Do not reveal the Sample Response table until the submission window closes — this is a private assessment, not a worked discussion.
Amplify · 2 inputs

Order per problem: Text component (problem text + Example: 1.5051) → Math Response (named answer1 / answer2) → Note with CL in </>.

Both problems use numericValue with ±0.0005 windows centered on the four-decimal-place answer.

Problem 1: Solve 10^x = 32. Express x as a logarithm, then compute to 4 decimal places.
content: when answer1.submitted and answer1.numericValue > 1.5046 and answer1.numericValue < 1.5056 "Correct — x = log 32 ≈ 1.5051. Bracket check: 10^1 = 10 and 10^2 = 100, so log 32 is between 1 and 2 — closer to halfway. Calculator confirms 1.5051." when answer1.submitted and answer1.numericValue > 1.49 and answer1.numericValue < 1.52 "Very close — keep four decimal places. log(32) on the calculator reads 1.5051." when answer1.submitted and answer1.numericValue > 0.99 and answer1.numericValue < 2.01 "You're in the right bracket — log 32 is between 1 and 2. Type log(32) on the calculator and round to four decimal places. Example: 1.5051" when answer1.submitted and isBlank(answer1.latex) "Express 10^x = 32 as x = log 32, then compute. Example: 1.5051" when answer1.submitted "Take another look — rewrite 10^x = 32 in log form first: x = log 32. Then type log(32) into the calculator and round to four decimal places." otherwise ""
Problem 2: Solve 10^x = 0.25. Express x as a logarithm, then compute to 4 decimal places.
content: when answer2.submitted and answer2.numericValue > -0.6026 and answer2.numericValue < -0.6016 "Correct — x = log 0.25 ≈ -0.6021. The answer is negative because 0.25 is between 0 and 1, and the warm-up sign rule says logs of numbers under 1 are negative." when answer2.submitted and answer2.numericValue > 0.6016 and answer2.numericValue < 0.6026 "Almost — check the sign. 0.25 is less than 1, so its log is negative. Try -0.6021." when answer2.submitted and answer2.numericValue > -0.62 and answer2.numericValue < -0.58 "You're very close — keep four decimal places. log(0.25) on the calculator reads -0.6021." when answer2.submitted and isBlank(answer2.latex) "Express 10^x = 0.25 as x = log 0.25, then compute. Example: -0.6021" when answer2.submitted "Rewrite 10^x = 0.25 as x = log 0.25. Then type log(0.25) into the calculator and watch the sign — 0.25 is under 1, so the log is negative." otherwise ""
Lesson Summary ~2 min

Lesson Summary

Sometimes a logarithm has an exact integer value. For example, \(\log 1000 = 3\) because \(10^3 = 1000\), and \(\log_2 32 = 5\) because \(2^5 = 32\).
Often the value is not an integer — but we can still estimate. To estimate \(\log 980\): the powers of 10 that bracket 980 are \(10^2 = 100\) and \(10^3 = 1000\), so \(\log 980\) is between 2 and 3. Because 980 is much closer to 1000 than to 100, \(\log 980\) is much closer to 3 — about 2.9.
The calculator confirms: log(980) \(\approx\) 2.991. The estimate told us what to expect; the calculator gave the precision.
Lesson Summary A logarithm answers the question what exponent makes this true? When the answer is not a whole number, bracket the argument between consecutive powers of 10 to estimate, then use the log button to check.
Lesson Summary slide. Optional read-aloud or self-read at the end of class. The two-step rhythm — estimate first, then verify — is the habit we want students to carry forward; the calculator is a confirmation tool, not a first move. Preview tomorrow in one sentence: “Tomorrow we will see what happens when the base is not 10 — the calculator will still help, but we will need log(x)/log(b), the change-of-base form.” Do not assess change of base today; just plant the seed.