NumeriveMATH
CurriculumFamily workspaceStudentMy characterAchievementsContact
← Intermediate Algebra levels

Intermediate Algebra / LEVEL 4 · DIFFICULTY 4/5

Functions and Identities

Simplify logarithmic, absolute-value and symmetric relationships.

3 stages · 24 practice problems · two 6-question assessment forms

Choose an island to read its lesson.

  1. MINI QUEST IA 4.1Logarithmic StructureRead the lesson
  2. MINI QUEST IA 4.2Absolute Value as DistanceRead the lesson
  3. MINI QUEST IA 4.3Symmetric ExpressionsRead the lesson
  4. LEVEL CHECKCastle challenge

    6 questions across this level.

    Sign in for the level check →

STAGE IA 4.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Logarithmic Structure

Useful preparation: Bounding Positive Expressions

Goal: Understand and apply logarithmic structure.

Before you begin: Bounding Positive Expressions

Understand the idea

A logarithm names an exponent. Rewriting positive arguments as powers of the same base makes the exponent relationships visible and avoids approximate computation.

log_b(b^k)=k, b>0 and b≠1

Choose and carry out a method

Identify the power of the base in each argument. Replace each logarithm with that exponent, then apply the operation requested outside the logs.

Check the reasoning

Logarithm arguments must be positive and the base must be positive and unequal to one. Rules for products do not become rules for sums.

WORKED EXAMPLE 1

Evaluate log_3(9) + log_3(3).

  1. A logarithm is the exponent needed to obtain its argument.
  2. 9=3^2 and 3=3¹.
  3. The scores add to 2+1=3. Both arguments and the base satisfy the logarithm restrictions.

3

WORKED EXAMPLE 2

Evaluate log_4(16) + log_4(4).

  1. A logarithm is the exponent needed to obtain its argument.
  2. 16=4^2 and 4=4¹.
  3. The scores add to 2+1=3. Both arguments and the base satisfy the logarithm restrictions.

3

Common pitfalls

Possible mix-up: log(a+b)=log a+log b.

The addition of logs corresponds to multiplying arguments.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Explain a logarithm by asking a question about an exponent.

Preview the eight practice prompts
  1. Evaluate log_3(27) + log_3(3).
  2. Evaluate log_4(64) + log_4(4).
  3. Evaluate log_2(16) + log_2(2).
  4. Evaluate log_3(81) + log_3(3).
  5. Evaluate log_4(256) + log_4(4).
  6. Evaluate log_2(32) + log_2(2).
  7. A scale assigns score log_3(v) to positive value v. Two readings are 243 and 3; what is the sum of their scores? New context
  8. A scale assigns score log_4(v) to positive value v. Two readings are 1024 and 4; what is the sum of their scores? New context
Open stage IA 4.1 in the student workspace →

STAGE IA 4.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Absolute Value as Distance

Useful preparation: Logarithmic Structure

Goal: Understand and apply absolute value as distance.

Before you begin: Logarithmic Structure

Understand the idea

The equation |x−h|=r places x at distance r from h on a number line. For positive r, there are two symmetric positions, one on each side.

|x−h|=r ⇒ x=h±r for r≥0

Choose and carry out a method

Write x−h=r or x−h=−r. Solve both linear equations and combine the values in the way the question requests.

Check the reasoning

A negative radius would give no real solutions. With radius zero, the two positions coincide and must not be counted twice.

WORKED EXAMPLE 1

The equation |x-3|=6 has two real solutions. Find their sum.

  1. Absolute value measures distance, so there is one solution on each side of the center.
  2. x=3-6 or x=3+6.
  3. The sum is 2·3=6; the distance terms cancel.

6

WORKED EXAMPLE 2

The equation |x-4|=7 has two real solutions. Find their sum.

  1. Absolute value measures distance, so there is one solution on each side of the center.
  2. x=4-7 or x=4+7.
  3. The sum is 2·4=8; the distance terms cancel.

8

Common pitfalls

Possible mix-up: Only x=h+r is possible.

A positive distance allows positions on both sides of h.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why is the sum of the two solutions independent of the radius?

Preview the eight practice prompts
  1. The equation |x-6|=9 has two real solutions. Find their sum.
  2. The equation |x-7|=10 has two real solutions. Find their sum.
  3. The equation |x-8|=11 has two real solutions. Find their sum.
  4. The equation |x-9|=12 has two real solutions. Find their sum.
  5. The equation |x-10|=13 has two real solutions. Find their sum.
  6. The equation |x-11|=14 has two real solutions. Find their sum.
  7. Two markers on a number line are each 15 units from position 12. What is the sum of their coordinates? New context
  8. Two markers on a number line are each 16 units from position 13. What is the sum of their coordinates? New context
Open stage IA 4.2 in the student workspace →

STAGE IA 4.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Symmetric Expressions

Useful preparation: Absolute Value as Distance

Goal: Understand and apply symmetric expressions.

Before you begin: Absolute Value as Distance

Understand the idea

Some expressions depend only on the sum and product of two numbers, so finding the individual numbers creates unnecessary work. Squaring the sum exposes the needed cross term.

r²+s²=(r+s)²−2rs

Choose and carry out a method

Expand (r+s)²=r²+2rs+s² and rearrange to obtain r²+s². Substitute the known sum and product directly.

Check the reasoning

This identity holds for real or complex values. Keep the coefficient two on the product term.

WORKED EXAMPLE 1

Numbers r and s satisfy r+s=8 and rs=3. Find r²+s².

  1. Expand the square of the sum and subtract the cross-term.
  2. r²+s²=(r+s)²-2rs=8²-2·3.
  3. The requested value is 58; individual roots are unnecessary.

58

WORKED EXAMPLE 2

Numbers r and s satisfy r+s=9 and rs=4. Find r²+s².

  1. Expand the square of the sum and subtract the cross-term.
  2. r²+s²=(r+s)²-2rs=9²-2·4.
  3. The requested value is 73; individual roots are unnecessary.

73

Common pitfalls

Possible mix-up: The square of a sum equals the sum of squares.

The expansion includes the cross term 2rs.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Find an expression for (r−s)² using the same two pieces of information.

Preview the eight practice prompts
  1. Numbers r and s satisfy r+s=11 and rs=6. Find r²+s².
  2. Numbers r and s satisfy r+s=12 and rs=7. Find r²+s².
  3. Numbers r and s satisfy r+s=13 and rs=8. Find r²+s².
  4. Numbers r and s satisfy r+s=14 and rs=9. Find r²+s².
  5. Numbers r and s satisfy r+s=15 and rs=10. Find r²+s².
  6. Numbers r and s satisfy r+s=16 and rs=11. Find r²+s².
  7. Two positive lengths have sum 17 and product 12. A model needs the sum of their squares. Find that value without solving for the individual lengths. New context
  8. Two positive lengths have sum 18 and product 13. A model needs the sum of their squares. Find that value without solving for the individual lengths. New context
Open stage IA 4.3 in the student workspace →
Browse another level →