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Topic 1.7Math AA SL SL30 flashcards

Exponent & log laws

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Card 1 of 301.7.1
1.7.1
Question

State the three index laws for the same base.

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All Flashcards in Topic 1.7

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1.7.110 cards

Card 1formula
Question

State the three index laws for the same base.

Answer

aᵐ × aⁿ = aᵐ⁺ⁿ (multiply→add); aᵐ ÷ aⁿ = aᵐ⁻ⁿ (divide→subtract); (aᵐ)ⁿ = aᵐⁿ (power of a power→multiply).

Card 2concept
Question

What is a⁰?

Answer

a⁰ = 1 for any a ≠ 0. Example: 7⁰ = 1.

Card 3formula
Question

What does a negative exponent mean?

Answer

A reciprocal: a⁻ⁿ = 1/aⁿ. Example: 2⁻³ = 1/8.

Card 4formula
Question

What does a fractional exponent mean?

Answer

A root: a^(1/n) = ⁿ√a, and a^(m/n) = (ⁿ√a)ᵐ. Example: 8^(2/3) = (∛8)² = 4.

Card 5concept
Question

Evaluate 27^(2/3).

Answer

Cube root first (∛27 = 3), then square: 3² = 9.

Card 6concept
Question

Write 1/√x as a power of x.

Answer

√x = x^(1/2), and the reciprocal flips the sign: 1/√x = x^(−1/2).

Card 7concept
Question

Given a^(2/3) = 4, find a.

Answer

Raise both sides to the reciprocal 3/2: a = 4^(3/2) = (√4)³ = 8.

Card 8concept
Question

Can you combine 2³ × 3² with the index laws?

Answer

No — the laws need the SAME base. 2³ × 3² = 8 × 9 = 72 must be done directly.

Card 9concept
Question

How do you solve an equation with aˣ and a²ˣ?

Answer

It is a quadratic in disguise: a²ˣ = (aˣ)², so substitute y = aˣ, solve the quadratic, then solve back for x.

Card 10concept
Question

In a quadratic-in-aˣ, why reject y ≤ 0?

Answer

Because y = aˣ and a power is always positive — only positive y can give a real x. Discard zero or negative roots.

1.7.210 cards

Card 11concept
Question

You see 2 log x + log y − log z. Which way do the log laws take you, and to what?

Answer

Coefficients go UP as powers, + becomes ×, − becomes ÷: 2 log x + log y − log z = log(x²y/z). To go the other way (one log → several), read the laws right-to-left. Spot whether they want 'a single log' or 'expanded'.

Card 12concept
Question

Can you split log(x + y)?

Answer

No — that is the #1 trap. The laws only act on products, quotients and powers, never on a sum.

Card 13concept
Question

What are log_a 1 and log_a a?

Answer

log_a 1 = 0 (since a⁰ = 1) and log_a a = 1 (since a¹ = a).

Card 14formula
Question

The power law — what does it do?

Answer

It brings an exponent down to the front as a coefficient: log_a(xᵐ) = m log_a x. Example: log 8 = log 2³ = 3 log 2.

Card 15concept
Question

Write ln 6 + 2 ln 3 − ln 2 as a single logarithm.

Answer

2 ln 3 = ln 9; then ln 6 + ln 9 − ln 2 = ln(6 × 9 ÷ 2) = ln 27.

Card 16concept
Question

Given log 2 = p and log 3 = q, write log 24 in terms of p and q.

Answer

24 = 2³ × 3, so log 24 = 3 log 2 + log 3 = 3p + q.

Card 17concept
Question

Your calculator only does log₁₀ and ln, but you need log₂ 50. What do you do?

Answer

Change of base: log₂ 50 = (log 50)/(log 2) ≈ 5.64 (any base b works: log_a x = (log_b x)/(log_b a)). Use it whenever the base is not 10 or e, or to combine logs of different bases.

Card 18concept
Question

Evaluate log₈ 32.

Answer

Change to base 2: log₂32 ÷ log₂8 = 5 ÷ 3 = 5/3.

Card 19concept
Question

Given log 2 = p and log 3 = q (base 10), write log₃ 8 in terms of p and q.

Answer

Change to base 10: log 8 ÷ log 3 = 3 log 2 ÷ log 3 = 3p/q.

Card 20concept
Question

Expand log₂(8x³).

Answer

Product then power law: log₂8 + log₂x³ = 3 + 3 log₂ x.

1.7.310 cards

Card 21concept
Question

How do you solve aˣ = b when the bases can be matched?

Answer

Write both sides as powers of the same base, then equate the exponents. E.g. 4ˣ = 8 → 2²ˣ = 2³ → 2x = 3 → x = 3/2.

Card 22concept
Question

How do you solve aˣ = b when the bases will not match?

Answer

Take logs of both sides; the power law brings x down: x log a = log b, so x = log b / log a.

Card 23concept
Question

Why does taking logs solve an exponential equation?

Answer

The power law: log aˣ = x log a turns the unknown exponent into a coefficient you can divide out.

Card 24concept
Question

Solve 5ˣ = 20 exactly.

Answer

x log 5 = log 20, so x = log 20 / log 5 = log₅ 20 (≈ 1.86).

Card 25formula
Question

How do you solve a log equation like log_a(expr) = c?

Answer

Convert to exponential form: expr = aᶜ, then solve. E.g. log₃(x − 1) = 2 → x − 1 = 9 → x = 10.

Card 26concept
Question

Solve ln(x² − 16) = 0.

Answer

e⁰ = x² − 16, so x² = 17 and x = ±√17 (both keep the argument positive).

Card 27concept
Question

Two logarithms in one equation — what is the first step?

Answer

Combine them into a single log with the product or quotient law, then convert to exponential form and solve.

Card 28concept
Question

Why must you check solutions of a log equation?

Answer

A logarithm needs a positive argument, so reject any solution that makes an argument ≤ 0.

Card 29concept
Question

Given log_k 81 = 4, find the base k.

Answer

k⁴ = 81, so k = ⁴√81 = 3 (take the positive root).

Card 30concept
Question

On Paper 2, how do you solve an exponential equation graphically?

Answer

Enter each side as Y₁ and Y₂, graph, then use 2nd → TRACE → 5: intersect. Check for more than one crossing.

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IB Math AA SL SL Topic 1.7 Flashcards | Exponent & log laws | Aimnova | Aimnova