g10-06-09 · Core topics
A decreasing base reverses the inequality
Does logₐx<logₐb always imply x<b?
1 MIN · PREDICT ON PAPER
Solve log₂x<log₂4 and log_(1/2)x<log_(1/2)4 before testing a value.
Discuss or write first. No typed answer is required to continue.
Solution: 0 < x < 4. Current input satisfies the strict inequality.
2 MIN · EXPLAIN AND REVISE
Which condition does your claim need?
Compare at least two cases. Write a reason, then check the explanation together.
Reveal explanation and conditions
For a>1, logₐx<logₐb gives 0<x<b and aˣ<aᵇ gives x<b. For 0<a<1, both functions decrease and both comparisons instead give x>b. Equality x=b is excluded.
Conditions: Require a>0,a≠1. Logarithmic arguments x,b must be positive; exponential inputs x,b may be any real numbers. Base 1 is excluded here.
Mathematical justification: Strict monotonicity gives equivalence with x<b for an increasing log and x>b for a decreasing log. The positive-argument restriction must be intersected with the result.
1 MIN · EXIT CHECK
Solve log_(1/2)x<log_(1/2)4.
Check after answering
x>4.
Teacher guidance and extension
Before: Real-number operations and reading coordinates; write a prediction on paper.
Ask: Keep the same test input while switching only the base from 2 to 1/2.
Watch for: Taking logarithms always preserves the inequality direction.
Challenge: State the corresponding rule for aˣ<aᵇ and compare the domain restrictions.
Suggested timing: predict 1 min, compare 3 min, explain 2 min, exit check 1 min. Observations can suggest a claim; they do not prove it for every input.
Optional local notebook
Paper responses are enough. Drafts remain in this tab; choose local saving to include them in the existing complete backup.
Complete backup and restore