HSF.LE.A.4: Solving Exponential Equations with Logarithms
In plain English: HSF.LE.A.4 is the Common Core functions standard that asks students to solve exponential equations of the form ab^(ct) = d by writing the solution as a logarithm, t = log_b(d/a)/c, when the base b is 2, 10 or e, and then to evaluate that logarithm with a calculator. It is usually taught in Algebra II.
For exponential models, express as a logarithm the solution to abᶜᵗ = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology.
Common Core State Standards for Mathematics · Domain: Linear, Quadratic, and Exponential Models (LE) · Cluster: Construct and compare linear, quadratic, and exponential models and solve problems Also written as HSF-LE.A.4 or F-LE.4 · Official standard
Students solve exponential equations of the form abct = d, where the base b is 2, 10 or e. They isolate the power, rewrite the exponent as a logarithm and give the exact answer t = logb(d/a)/c. Then they evaluate that logarithm with a calculator or graphing tool and check the decimal in the original model.
Every example comes from an exponential model: doubling cells, continuous interest, a drug leaving the body, the pH of rain. Students learn which key goes with which base, see on a graph that the logarithm is the input where the curve reaches the target value, and recognize when an equation has no solution because d/a is not positive.
Learning Objectives
By the end of this lesson, students will be able to:
Isolate the power in abct = d and express the solution as t = logb(d/a)/c
Solve exponential equations with base 2, base 10 and base e, including models with a negative c
Evaluate log, ln and log2 expressions with a calculator or graphing tool and round sensibly
Check a logarithmic solution by substitution or on a graph, and interpret it in the context of the model
Explain why abct = d has no solution when d/a is zero or negative
Prior Knowledge Required
Students should already be comfortable with:
Properties of integer exponents 8.EE.A.1
Writing exponential models such as a · 2t/k or Pert from a context HSF.LE.A.2
Solving simple equations such as 2t = 16 by inspection
Using a scientific or graphing calculator, including the order of operations for entries with parentheses
Post three equations and ask students to find t without a calculator, or to say as precisely as they can where t lies:
Warm-Up Prompt
"Find t: (1) 2t = 8, (2) 10t = 1,000, (3) 2t = 12. For (3), between which two whole numbers is t? How could you find it exactly?"
Students get t = 3 for both (1) and (2). For (3), a table of powers of 2 shows 23 = 8 and 24 = 16, so t is between 3 and 4. Tell students that this exponent has a name: t = log2 12, "the power of 2 that gives 12." A calculator gives about 3.585, and 23.585 ≈ 12. The rest of the lesson turns this idea into a method for every equation of the form abct = d.
Direct Instruction20-25 minutes
The definition. For a base b > 0 (b ≠ 1), logb y = x means bx = y. The standard uses three bases: 2 (doubling and halving), 10 (the LOG key, written log with no base) and e (the LN key, written ln). Then show the four steps in Diagram 2 for solving abct = d:
Isolate the power: divide both sides by a to get bct = d/a. Do not take a logarithm of a · bct as a whole.
Check the sign: if d/a ≤ 0 there is no solution, because every power of a positive base is positive.
Write the exponent as a logarithm: ct = logb(d/a).
Divide by c: t = logb(d/a) / c. This is the exact answer the standard asks for.
Evaluate with technology and round only at the end. Check by substituting the decimal back into the model.
On calculators, LOG is base 10 and LN is base e. For base 2, use a log-base template if the calculator has one (graphing tools such as Desmos accept log2 directly), or enter log(x) ÷ log(2). Work the examples below, one for each base, and name the base before starting.
Base 2, doubling model
A culture starts with 300 cells and doubles every 4 hours: N = 300 · 2t/4. When does it reach 5,000 cells?
Equation: 2t/4 = 50/3, so t = 4 log2(50/3) ≈ 16.24 hours
Base 10, growth model
A website has V = 2,500 · 100.15t visitors per month, t months after launch. When does it reach 40,000 visitors per month?
Equation: 100.15t = 16, so t = (log 16)/0.15 ≈ 8.03 months
Base e, continuous growth
$1,200 earns 4.5% compounded continuously: A = 1200e0.045t. When is the balance $2,000?
Equation: e0.045t = 5/3, so t = ln(5/3)/0.045 ≈ 11.35 years
Base e, decay with negative c
A 250 mg dose leaves the body as A = 250e-0.35t, with t in hours. When is 40 mg left?
Equation: e-0.35t = 0.16, so t = ln(0.16)/(-0.35) ≈ 5.24 hours
Base 10, no context
Express the solution of 6 · 102t = 42 as a logarithm and evaluate it.
Equation: 102t = 7, so t = (log 7)/2 ≈ 0.4225
After the fourth example, point out that c can be negative, and the negative log divided by a negative c gives a positive time. After the fifth, ask what would happen with 3 · 102t = -12: dividing gives 102t = -4, which no real t satisfies. Use Diagram 1 to show that the logarithm in the first example is exactly the t-coordinate where the curve meets the line N = 5,000.
Guided Practice15 minutes
Pairs solve three equations, one per base, writing the exact logarithm before they touch the calculator: 7 · 2t = 91 (t = log2 13 ≈ 3.70), 4 · 100.5t = 900 (t = 2 log 225 ≈ 4.70) and 50e0.8t = 20 (t = ln(0.4)/0.8 ≈ -1.15). After each one, a pair shows its calculator entry. Discuss the third: a negative t means 1.15 time units before t = 0, which is fine in a model that runs both ways. Listen for these errors: taking the log before dividing by a, pressing LOG for a base-e equation, forgetting to divide by c, and entering log 13 ÷ 2 instead of log 13 ÷ log 2.
Independent Practice15 minutes
Students solve four equations on their own, giving an exact logarithm and a decimal for each: 5 · 22t = 60 (t = (log2 12)/2 ≈ 1.79), 8 · 103t = 200 (t = (log 25)/3 ≈ 0.466), 12e-0.2t = 3 (t = ln(0.25)/(-0.2) ≈ 6.93) and 3e2t = 45 (t = (ln 15)/2 ≈ 1.35). For each, students substitute their decimal back into the left side and confirm they get d, within rounding.
Closure5-10 minutes
Exit ticket: (1) Solve 9 · 20.5t = 72 and explain why no calculator is needed. (20.5t = 8, so t = 2 log2 8 = 6.) (2) Express the solution of 2e0.1t = 11 as a logarithm and evaluate it. (t = 10 ln 5.5 ≈ 17.05.) (3) Which calculator key does each base use: 2, 10, e?
Differentiation Strategies
For Struggling Students
Give a table of powers of 2 and 10 so students can first bracket each answer between two whole numbers, then compare it with the calculator value
Use a three-column organizer: isolated power, log form, calculator entry. Students fill the first two columns before touching the calculator
Start with equations where a = 1 and c = 1, such as 10t = 60, before adding a and c
For Advanced Students
Ask students to solve 5 · 2t = 3 · 10t by taking the log of both sides, and explain why this equation is not of the form abct = d (extension beyond the standard)
Ask students to show that log(x) ÷ log(2) and ln(x) ÷ ln(2) always give the same number, and explain why using the definition of a logarithm
Ask students to find the doubling time of Pert in general and compare (ln 2)/r with the "rule of 70" estimate 70/(100r)
Assessment Guidance
What to Look For
Check that students write the exact logarithm, such as t = 4 log2(50/3), before giving a decimal: the standard asks for both. Watch the order of steps: the power must be isolated before the logarithm is written, and c is divided out after it. When a student reports a decimal, ask for the calculator entry, because a correct setup with the wrong key (LOG for base e) is a common error. In context problems, look for units and a sentence of interpretation, and for students noticing when a negative answer or no answer makes sense.
02
Classroom Activities
3 Activities
1
Three Bases Card Match
20 minGroups of 3-4
Groups get 27 cards: 9 equation cards, 9 exact-logarithm cards and 9 decimal cards, three equations for each base. They build 9 rows of three matching cards. Matching exact answers first, and decimals second, keeps attention on the logarithm form the standard asks for.
Groups first match each equation card with its exact-logarithm card, writing the "isolate the power" step on a mini whiteboard
Then one student evaluates each logarithm on a calculator and the group matches the decimal card
Two equations have whole-number answers. Groups find them and explain why no calculator is needed
Groups check two rows of another group by substituting the decimal back into the equation
Challenge Variation
Give each group three blank cards and ask for one new equation per base whose solution is between 2 and 3. Groups trade and solve.
2
Estimate First, Then Compute
20 minPairs
For three models, pairs first bracket the answer between two numbers using known powers, then write the exact logarithm and evaluate it. The estimate is a check on the technology step and catches wrong keys.
Models
Base 2: a sourdough starter grows as V = 150 · 2t/6 mL. When is it 1,000 mL? (2t/6 = 20/3 is between 4 and 8, so t is between 12 and 18. Exact: t = 6 log2(20/3) ≈ 16.42 hours)
Base 10: a sound level of L decibels is 10L/10 times the quietest audible sound. What level is 3,000,000 times it? (Between 106 and 107, so L is between 60 and 70. Exact: L = 10 log 3,000,000 ≈ 64.77 dB, about the level of a conversation)
Base e: $900 at 4.2% compounded continuously, A = 900e0.042t. When is A = $2,000? (e0.042t ≈ 2.22 is between 1 and e, so 0.042t is between 0 and 1. Exact: t = ln(20/9)/0.042 ≈ 19.01 years)
Procedure
Partner A writes the bracket and Partner B writes the exact logarithm; they swap for the next model
Both evaluate the logarithm and circle the answer only if it falls inside the bracket
Pairs write one sentence for each model interpreting the answer with units
Discussion Questions
Your partner got 7.13 for the sourdough model. Without redoing it, how do you know it is wrong?
For the sound model, a = 1 and c = 1/10. Which step of the method disappears when a = 1?
3
Check It on a Graph
15 minIndividual, then pairs
Students solve three equations with logarithms, then graph both sides in a graphing tool and read the intersection. Seeing the same number twice connects the logarithm to the graph of the model, as in Diagram 1.
Equations
40 · 2x/5 = 250 (x = 5 log2 6.25 ≈ 13.22)
7 · 100.2x = 350 (x = 5 log 50 ≈ 8.49)
600e-0.25x = 75 (x = 4 ln 8 ≈ 8.32)
Procedure
Solve each equation on paper: exact logarithm, then decimal
Graph y = left side and y = right side and tap the intersection point. Record its x-coordinate next to your decimal
With a partner, graph y = 40 · 2x/5 and y = -20. Explain why the graphs never meet and what that says about the equation
Modification for Distance Learning
Share one graphing link per pair. Each student adds the lines for one equation in a different color and posts a screenshot of the intersection next to the logarithm.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Logarithm Is Where the Curve Meets the Target
The model N = 300 · 2t/4 drawn to scale for 0 ≤ t ≤ 18 hours, with the target line N = 5,000. The curve crosses the line at t = 4 log2(50/3) ≈ 16.24 hours, the same value the logarithm gives.
Diagram 2: Four Steps and Three Calculator Keys
The general method for abct = d, followed by the technology step for each base in the standard. The sample values come from the worked examples: log 16 (base 10), ln(5/3) (base e) and log2(50/3) (base 2).
04
Homework Assignment
~30 min
HSF.LE.A.4 Homework: Exponential Equations and Logarithms
Directions: For every problem, isolate the power, write the exact solution as a logarithm, then evaluate it with technology to the nearest hundredth. Write the calculator entry you used. In Part 2, answer in a sentence with units.
Part 1: Writing the Logarithm (Problems 1-3)
A new app has U = 1,500 · 2t/3 users, t months after launch. Write the time when it reaches 20,000 users as a base-2 logarithm, then evaluate it.
Solve 25 · 100.4t = 800.
Solve 80e-0.05t = 30. Explain why the answer is positive even though c is negative.
Part 2: Models in Context (Problems 4-6)
$3,000 is deposited at 3.8% interest compounded continuously, so A = 3000e0.038t. When will the account hold $4,500?
In an invented model of a lake, light intensity at depth d meters is I = 900 · 10-0.12d units. At what depth is the intensity 50 units?
Iodine-131 has a half-life of 8 days, so 120 mg decays to A = 120 · 2-t/8 mg. (a) When is 15 mg left? Solve without a calculator. (b) When is 5 mg left? (c) Explain why 120 · 2-t/8 = 0 has no solution, and what that means for the iodine.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Isolating the Power
Divides by a before writing any logarithm
Correct idea with one arithmetic error
Takes a log of the whole left side or skips the step
Exact Logarithm
Correct base and argument, divided by c
Correct log with c missing or misplaced
No logarithm written
Technology
Correct key or entry for each base, decimal correct
Correct entry, rounding error
Wrong key for the base
Interpretation
Answers with units, explains no-solution case
Units or explanation missing once
No interpretation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
You need a calculator with LOG and LN keys or a graphing tool. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which expression is the exact solution of 2t = 45?
Answer: C
By definition, 2t = 45 means t = log2 45 (about 5.49). Choice A uses base 10, which answers 10t = 45. Choice B divides by the base, as if the equation were 2t = 45. Choice D swaps the base and the argument.
Question 2 of 20 · Multiple Choice
Solve 4 · 10t = 36 exactly.
Answer: B
Divide by 4 first: 10t = 9, so t = log 9 (about 0.954). Choice A subtracts 4 instead of dividing. Choice C takes the log before isolating the power; log(4 · 10t) is not 4t. Choice D treats 4 as the base.
Question 3 of 20 · Multiple Choice
Solve e0.2t = 7 exactly.
Answer: C
Write the power as a logarithm: 0.2t = ln 7, so t = (ln 7)/0.2 = 5 ln 7 (about 9.73). Choice A multiplies by 0.2 instead of dividing by it. Choices B and D divide or multiply 7 by 5 inside the logarithm, which changes the equation.
Question 4 of 20 · Multiple Choice
A population is modeled by 6 · 20.5t = 90. To the nearest hundredth, what is t?
Answer: D
Divide by 6: 20.5t = 15. Then 0.5t = log2 15, so t = 2 log2 15 ≈ 2(3.907) ≈ 7.81. Choice A forgets to divide by c = 0.5. Choice B uses the LOG key (base 10) instead of base 2. Choice C skips dividing 90 by 6.
Question 5 of 20 · Multiple Choice
Use technology to evaluate log 250 to the nearest thousandth.
Answer: A
The LOG key is base 10: log 250 ≈ 2.398, and 102.398 ≈ 250. Choice B is ln 250 (the LN key, base e). Choice C is log2 250. Choice D divides 250 by 10.
Question 6 of 20 · Multiple Choice
A calculator has only LOG and LN keys. Which entry gives log2 20?
Answer: C
log2 20 is the power of 2 that gives 20, and log(20) ÷ log(2) ≈ 4.322 has that property: 24.322 ≈ 20. ln(20) ÷ ln(2) gives the same value. Choice A is log 10 = 1. Choice B is upside down: it gives about 0.231, the reciprocal.
Question 7 of 20 · Multiple Choice
$5,000 is invested at 6% interest compounded continuously, so A = 5000e0.06t. After how many years, to the nearest hundredth, will the balance be $8,000?
Answer: C
Divide by 5000: e0.06t = 1.6, so 0.06t = ln 1.6 and t = (ln 1.6)/0.06 ≈ 7.83 years. Choice A uses base 10 (LOG) instead of base e. Choice B stops at ln 1.6 without dividing by 0.06. Choice D subtracts 5000 from 8000 instead of dividing.
Question 8 of 20 · Multiple Choice
A medicine level is modeled by 90e-0.4t = 18. Which is the exact solution?
Answer: A
e-0.4t = 18/90 = 0.2, so -0.4t = ln 0.2 and t = ln(0.2)/(-0.4) ≈ 4.02 hours. Choice B drops the negative sign on c and gives a negative time. Choice C multiplies by c instead of dividing. Choice D subtracts 90 instead of dividing by it.
Question 9 of 20 · Multiple Choice
The hydrogen ion concentration of a rainwater sample satisfies 10-p = 0.000032, where p is the pH. What is the pH, to the nearest hundredth?
Answer: B
Take the base-10 log: -p = log(0.000032) ≈ -4.49, so p ≈ 4.49 (acidic, as rain often is). Choice A forgets to divide by c = -1. Choice C uses ln instead of log. Choice D is off by one, as if the concentration were 0.0000032.
Question 10 of 20 · Multiple Choice
Which equation has no real solution?
Answer: D
In choice D, 2t = -7, and a power of 2 is always positive, so log2(-7) is undefined and there is no solution. Choices A and B both give 2t = 4 (t = 2). Choice C gives 2-t = 4, so t = -2: a negative c is allowed.
Question 11 of 20 · Multiple Choice
A video has V = 800 · 2t/2 views after t days. After how many days, to the nearest hundredth, does it reach 100,000 views?
Answer: A
Divide by 800: 2t/2 = 125. Then t/2 = log2 125, so t = 2 log2 125 ≈ 13.93 days. Choice B forgets to multiply by 2. Choice C uses base 10. Choice D doubles 125 instead of using a logarithm.
Question 12 of 20 · Multiple Choice
Use technology to evaluate ln 0.35 to the nearest thousandth.
Answer: D
ln 0.35 ≈ -1.050, and e-1.050 ≈ 0.35. The value is negative because 0.35 < 1. Choice A drops the sign. Choice B is log 0.35 (base 10), and choice C is log2 0.35.
Question 13 of 20 · Multiple Choice
Which step correctly rewrites 103t = 500?
Answer: A
If 103t = 500, the exponent 3t is the base-10 log of 500, so 3t = log 500 and t = (log 500)/3 ≈ 0.900. Choice B divides inside the log instead of dividing the log by 3. Choice C uses the wrong base. Choice D confuses a power with a logarithm.
Question 14 of 20 · Multiple Choice
A 5 mg dose of a drug has a half-life of 6 hours, so the amount left is 5 · 2-t/6 mg. When, to the nearest hundredth of an hour, is 2 mg left?
Answer: C
2-t/6 = 2/5 = 0.4, so -t/6 = log2 0.4 and t = -6 log2 0.4 ≈ 7.93 hours. Check: between one half-life (2.5 mg at 6 hours) and two (1.25 mg at 12 hours). Choice A forgets the negative sign in c. Choice B uses base 10. Choice D stops at log2 2.5 and does not multiply by 6.
Question 15 of 20 · Short Answer
Solve 12 · 100.25t = 300. Give the exact solution as a logarithm and a decimal to the nearest hundredth.
Divide by 12: 100.25t = 25. Then 0.25t = log 25, so t = 4 log 25 ≈ 5.59. Check: 12 · 101.398 ≈ 12 · 25 = 300.
Question 16 of 20 · Short Answer
A town of 42,000 people grows continuously: P = 42,000e0.012t, with t in years. Write the time when the population reaches 50,000 as a logarithm, then evaluate it.
e0.012t = 50,000/42,000 = 25/21. So 0.012t = ln(25/21) and t = ln(25/21)/0.012 ≈ 14.53 years, during the 15th year.
Question 17 of 20 · Short Answer
Solve 3 · 24t = 51 exactly, then evaluate with technology.
24t = 17, so 4t = log2 17 and t = (log2 17)/4 ≈ 1.02. With only LOG: (log 17 ÷ log 2) ÷ 4 ≈ 4.087 ÷ 4.
Question 18 of 20 · Short Answer
Evaluate log2 1000 with technology, to the nearest thousandth. Explain why your answer should be between 9 and 10.
log2 1000 ≈ 9.966. Since 29 = 512 < 1000 < 1024 = 210, the exponent must be between 9 and 10, and close to 10 because 1000 is close to 1024.
Question 19 of 20 · Short Answer
A $2,000 deposit earns 5% interest compounded continuously: A = 2000e0.05t. Write the doubling time as a logarithm and evaluate it.
Doubling means A = 4000, so e0.05t = 2, 0.05t = ln 2 and t = (ln 2)/0.05 ≈ 13.86 years. The starting amount cancels, so any deposit at this rate doubles in the same time.
Question 20 of 20 · Short Answer
Solve 0.5 · 10-0.3t = 0.02. Give the exact answer and a decimal to the nearest hundredth.
10-0.3t = 0.02/0.5 = 0.04. So -0.3t = log 0.04 and t = log(0.04)/(-0.3) ≈ 4.66. Both log 0.04 ≈ -1.398 and c are negative, so t is positive.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.LE.A.4 mean?
It means students can solve an exponential equation like 300 · 2t/4 = 5,000 by writing the answer as a logarithm and then finding its value with a calculator. The standard limits the base to 2, 10 or e, the three bases that have calculator keys or common models: doubling and halving, powers of ten, and continuous growth.
Is HSF.LE.A.4 Algebra 1 or Algebra 2?
It is usually taught in Algebra II. In Algebra I, students write and graph exponential models (HSF.LE.A.2) and estimate solutions from graphs or tables. Logarithms are introduced in Algebra II, where this standard gives the first exact way to solve for the exponent.
How do you solve abᶜᵗ = d with a logarithm?
Divide by a, write the exponent as a logarithm, then divide by c: t = logb(d/a)/c. For example, 6 · 102t = 42 gives 102t = 7, so 2t = log 7 and t = (log 7)/2 ≈ 0.42.
Which calculator button do I use for log base 2?
Use a log-base template if your calculator has one, or type log2 in a graphing tool such as Desmos. Otherwise enter log(x) ÷ log(2), or ln(x) ÷ ln(2); both give the same value. The LOG key alone is base 10, and LN is base e.
Why do I have to divide by a before taking the logarithm?
Because the logarithm undoes only the power bct, not the product a · bct. In 5 · 10t = 150, writing t = log 150 ignores the 5. Dividing first gives 10t = 30 and t = log 30 ≈ 1.477, which checks: 5 · 101.477 ≈ 150.
What are common mistakes on these problems?
A common one is pressing LOG for an equation with base e, or LN for base 2 without dividing by ln 2. Others are forgetting to divide by c at the end, dividing inside the logarithm (log(80/4) instead of (log 80)/4), and rounding the logarithm early, which can move the final answer by several hundredths.
Can an exponential equation have no solution?
Yes. If d/a is zero or negative, there is no solution, because bct is positive for every t. For example, 2 · 10t = -6 gives 10t = -3, and no power of 10 is negative. On a graph, the curve never reaches the target line.
Why is the answer sometimes negative?
A negative t means a time before the starting point of the model. For 50e0.8t = 20, t ≈ -1.15: the quantity was 20 about 1.15 time units before it was 50. Whether that makes sense depends on the context and on the time range where the model is valid.
Does this standard appear on the SAT?
Exponential functions and equations are part of the SAT Advanced Math domain. Many of those questions can be answered with the built-in graphing calculator by finding where a curve meets a horizontal line, the same idea as Diagram 1. Writing the exact logarithm is mainly assessed in Algebra II courses and state tests.
What comes after HSF.LE.A.4?
In a course with advanced (+) standards, HSF.BF.B.5 studies logarithms as the inverse of exponential functions and uses that relationship to solve problems. Students also meet logarithmic scales such as pH and decibels in science, and exponential equations with logarithms return in precalculus and calculus.
07
Related Standards
6 standards
These standards connect to HSF.LE.A.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSF.LE.A.2Prerequisite
Construct linear and exponential functions from a graph, description or two points
