Grade 11 Exponential & Logarithmic Functions Practice — Medium

Question 1 of 29Score 0/0Medium

Question 1 of 29: Expand: log₂(8x³/y²)

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Answer key for parents & teachers (29 questions)
  1. Expand: log₂(8x³/y²) — 3 + 3log₂x − 2log₂y. log₂(8x³/y²) = log₂8 + log₂x³ − log₂y² = 3 + 3log₂x − 2log₂y.
  2. Condense: 2log x + 3log y − log z — log(x²y³/z). 2log x = log x², 3log y = log y³, −log z = log(1/z). Combined: log(x²y³/z).
  3. b^(log_b x) = ? — x. b^(log_b x) = x. This is a fundamental inverse property of logarithms and exponentials.
  4. log_b(b^x) = ? — x. log_b(b^x) = x. Logarithm and exponential are inverse functions.
  5. Change of base formula: log_b(x) = ? — log(x)/log(b). Change of base: log_b(x) = log(x)/log(b) = ln(x)/ln(b). Use base 10 or base e.
  6. Using change of base, log₅(17) = ? — log(17)/log(5). log₅(17) = log(17)/log(5) = ln(17)/ln(5) ≈ 1.23/0.699 ≈ 1.76.
  7. Which is undefined? — log(0). log(0) is undefined — there is no exponent to which 10 can be raised to equal 0.
  8. log₄(1/16) = ? — −2. 1/16 = 4⁻², so log₄(1/16) = −2.
  9. Condense: log₃(9) + log₃(27) — 5. log₃(9) + log₃(27) = 2 + 3 = 5. (Or: log₃(9 · 27) = log₃(243) = log₃(3⁵) = 5.)
  10. log₂(√8) = ? — 1.5. √8 = 8^(1/2) = (2³)^(1/2) = 2^(3/2). log₂(2^(3/2)) = 3/2 = 1.5.
  11. Which property allows rewriting log(x³y²) as 3log x + 2log y? — Product rule and power rule. First apply the product rule: log(x³) + log(y²). Then the power rule: 3log x + 2log y.
  12. Solve: 4^x = 8 — x = 3/2. Write in base 2: 2^(2x) = 2³ → 2x = 3 → x = 3/2.
  13. Solve: 2^(3x) = 16 — x = 4/3. 2^(3x) = 2⁴ → 3x = 4 → x = 4/3.
  14. Why must you check for extraneous solutions in log equations? — Because log arguments must be positive. log(x) is undefined for x ≤ 0. Operations like squaring can produce solutions that make a log argument negative — these must be rejected.
  15. Solve: 3^(2x−1) = 27 — x = 2. 27 = 3³, so 2x − 1 = 3 → 2x = 4 → x = 2.
  16. Solve: log₂(x+3) = log₂(2x−1) — x = 4. If log_b(A) = log_b(B), then A = B. So x+3 = 2x−1 → 4 = x. Check: log₂(7) = log₂(7). ✓
  17. Solve: 10^(x−2) = 0.01 — x = 0. 0.01 = 10⁻² → x − 2 = −2 → x = 0.
  18. Solve: e^(2x) = 50 — x = ln(50)/2. 2x = ln(50) → x = ln(50)/2 ≈ 3.912/2 ≈ 1.956.
  19. Solve: ln(x) = −2 — x = e⁻². x = e^(−2) = 1/e² ≈ 0.135.
  20. Solve: 3·e^(2x) = 15 — x = ln(5)/2. e^(2x) = 5 → 2x = ln(5) → x = ln(5)/2.
  21. $1000 at 6% compounded continuously for 1 year yields: — $1061.84. A = 1000e^(0.06·1) = 1000e^0.06 ≈ 1000(1.06184) ≈ $1061.84.
  22. The half-life formula is t₁/₂ = ? — ln(2)/k. At half-life: A₀/2 = A₀e^(kt₁/₂) → 1/2 = e^(kt₁/₂) → ln(1/2) = kt₁/₂ → t₁/₂ = ln(2)/|k|.
  23. A population of 500 grows at r = 0.04 continuously. After 10 years: — ≈ 745. A = 500e^(0.04·10) = 500e^0.4 ≈ 500(1.4918) ≈ 745.
  24. Doubling time for continuous growth rate r is: — ln(2)/r. 2A₀ = A₀e^(rt_d) → 2 = e^(rt_d) → ln(2) = rt_d → t_d = ln(2)/r.
  25. Continuous compound vs. annual compounding at 5% for $1000 over 1 year, continuous gives: — More than annual. Continuous: $1000e^0.05 ≈ $1051.27. Annual: $1000(1.05) = $1050. Continuous gives slightly more.
  26. Newton's law of cooling: T(t) = T_env + (T₀ − T_env)e^(−kt). If an object cools from 100°C in 20°C room, which is T_env? — 20°C. T_env is the ambient/environment temperature — 20°C here. The object cools toward it.
  27. If a quantity doubles every 3 years, its continuous growth rate r is: — ln(2)/3. Doubling time t_d = ln(2)/r → 3 = ln(2)/r → r = ln(2)/3 ≈ 0.231 per year.
  28. The number e can be defined as lim(n→∞) of: — (1 + 1/n)^n. e = lim(n→∞)(1 + 1/n)^n ≈ 2.71828…
  29. Radioactive material with half-life 10 years: what fraction remains after 30 years? — 1/8. 30 years = 3 half-lives. (1/2)³ = 1/8.