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Solve for all positive x satisfying both equations.

Read the idea, work independently, then explain what changed.

TOPIC 01

2021 JM01

Both resulting x values are positive and satisfy the logarithm domain.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Solve for all positive x satisfying both equations.

log⁡4x=y−3,2(log⁡4x)2=4−y\log_4x=y-3,\quad2(\log_4x)^2=4-y

Official paper · jm01-2021 · I.10 · PDF 3

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A1/4 or 21/4\text{ or }2
  2. Option B1/2 or 41/2\text{ or }4
  3. Option C7/2 or 27/2\text{ or }2
  4. Option D1/4 or 7/21/4\text{ or }7/2
  5. Option E2 or 42\text{ or }4

Working and explanation

BUILD THE REASONING

Hint 1
Use u=log₄x and eliminate y.
Hint 2
Factor the resulting quadratic in u.
Worked solution
  1. The first equation gives y=u+3.

    2u2=1−u  ⟺  (2u−1)(u+1)=02u^2=1-u\iff(2u-1)(u+1)=0
  2. Convert both logarithm values back to x.

    u=1/2,−1⇒x=41/2=2, 4−1=1/4u=1/2,-1\Rightarrow x=4^{1/2}=2,\ 4^{-1}=1/4

A: x=1/4 or 2.

Checks and common pitfalls: Both resulting x values are positive and satisfy the logarithm domain.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The first equation gives y=u+3.
    2u2=1−u  ⟺  (2u−1)(u+1)=02u^2=1-u\iff(2u-1)(u+1)=0
  • Convert both logarithm values back to x.
    u=1/2,−1⇒x=41/2=2, 4−1=1/4u=1/2,-1\Rightarrow x=4^{1/2}=2,\ 4^{-1}=1/4

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Curriculum and source notes ↗