Precalculus and Trigonometry · 35 minutes

Suggested grades 11–12 · Unit 2: Exponential inverses and the unit circle · Lesson 1 of 2

Logarithms as Exponential Inverses

Translate exponential and logarithmic statements and solve a model on its real domain.

Mr. S Teaches · Original starter lesson · HS-PRE-03 · Grade range is planning guidance.

Watch, notice, discuss

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Graphing logarithmic functions | Exponential and logarithmic functions | Algebra II | Khan Academy

By Khan Academy

Compare the logarithmic graph with its exponential inverse and identify the positive-input restriction.

Open on YouTube: Graphing logarithmic functions | Exponential and logarithmic functions | Algebra II | Khan Academy (opens a new tab) ↗
After watching

How does the exponential-logarithmic inverse relationship solve the equation, and which inputs are allowed?

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Learning objective

Solve exponential equations using the inverse relationship and explain input restrictions.

Before you begin

  • Exponent laws
  • Algebra II logarithmic solutions

Gather your materials

  • Paper and pencil
  • Graph paper
  • Calculator for checking results

Language to know

logarithm base b
The exponent on positive base b≠1 that produces a positive number.
inverse relationship
Two operations reversing one another on a suitable domain.
domain restriction
A condition required for an expression to have the intended real value.

Understand the concept

The statements b^y=x and log_b(x)=y express the same relationship when b>0, b≠1 and x>0. Logarithm laws come from exponent laws, but a log of a sum cannot be split into a sum of logs. Solving must retain positivity conditions; an algebraically manipulated candidate may not belong to the original domain.

Teach it step by step

  1. Translate 3⁴=81 into log₃81=4 and back.
  2. Solve 3^t=20 as t=ln 20/ln 3 and check approximately by exponentiation.
  3. Solve log₂(x−1)=3, state x>1 first and test the candidate.
  4. Compare log₂(4+4) with log₂4+log₂4 to expose the false sum rule.

Make the thinking visible

Worked example or model response

log₂(x−1)=3 gives x−1=8 and x=9, which satisfies x>1. log₂8=3 but log₂4+log₂4=4, so the supposed sum rule fails.

Guided practice

Solve log₅(2x+1)=2 and state the real input condition.

Independent practice

1. Solve 2^(3t)=7 using logarithms. 2. Solve log₃(x−4)=0. 3. Explain why log₂(−8) has no real value.

Quick assessment

What must be true of the logarithm’s argument before using a real-valued logarithm?

Answer key & assessment criteria

Answers and teaching guidance

  1. Guided:2x+1>0, hence x>−1/2; 2x+1=25 gives x=12.
  2. Independent 1: t=ln 7/(3ln2)≈0.936. 2: x−4=1, so x=5. 3: No real power of 2 is negative.
  3. Quick check: The argument must be positive.

Look for: Require the learner to explain the interpretation and check, not only state a result. Distinguish an exact value from a rounded approximation and attend to the stated domain.

Make it work for your learner

More support

Use a paired exponential/logarithm statement table and highlight the argument.

Extension challenge

Create a logarithmic equation whose manipulated candidate violates the original domain and explain the rejection.

Verified references · Selected contributions

Standards connections

Subject-content standards

These records identify the part of a subject standard practiced here. Course placement is instructional guidance; the source’s category or band remains authoritative.

Contributes to Common Core standard

HSF.BF.B.5

Common Core · High school · Functions

(+) Additional preparation for advanced study

Use exponential/logarithmic inversion to solve and interpret equations.

Read the official standard ↗

Additional preparation (+). The lesson focuses on simple real-valued equations, not all logarithmic identities or models.

Verified 2026-11-04. Selected contribution, not full attainment.