37 exam-style questions on Laws of exponents (integer); introduction to logarithms — topic 1.5 of Number and Algebra in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
Simplify 3⁴ × 3⁻⁶, giving your answer as an integer or a fraction.
Answer: 1/9
When multiplying powers with the same base, add the exponents: 4 + (−6) = −2, so 3⁴ × 3⁻⁶ = 3⁻² = 1/9. A negative exponent means reciprocal, not a negative number.
Calculate the value of (2³)⁴ ÷ 2⁵.
Answer: 128
(2³)⁴ = 2¹² because a power of a power multiplies the exponents; dividing by 2⁵ subtracts exponents: 2¹²⁻⁵ = 2⁷ = 128.
Write 5⁻³ as a fraction.
Answer: 1/125
A negative exponent gives the reciprocal of the positive power: 5⁻³ = 1/5³ = 1/125. It does not make the value negative.
Which equation is equivalent to the statement log₅ 125 = 3?
Answer: 5³ = 125
By definition, log_a b = x means a^x = b: the base of the logarithm is the base of the power, and the logarithm itself is the exponent, so 5³ = 125.
Determine the value of x for which 2^x = 1/16.
Answer: −4
1/16 = 1/2⁴ = 2⁻⁴, so x = −4. Writing the fraction as a power of 2 with a negative exponent is the key step.
Simplify (2a³b⁻²)³.
Answer: 8a⁹/b⁶
Cube every factor: 2³ = 8, (a³)³ = a⁹ and (b⁻²)³ = b⁻⁶ = 1/b⁶. Multiplying the exponents (not adding) and cubing the coefficient are the common slips.
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