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Admissions tests / TMUA / Paper 1 / Exponentials and logarithms

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Exponentials and logarithms, set 2

Exponential graphs, the laws of logarithms, solving equations of the form a to the power x equals b, and using logarithms to linearise a relationship.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    The graph of y = a^x, where a > 0, passes through the points (2, 9) and (5, 243). Find a.

    1. A 234
    2. B 1/27
    3. C 3
    4. D 27
  2. 21 mark

    For the graph of y = (3/7)^x, which of the following statements is true?

    1. A The graph is decreasing for all real values of x
    2. B The graph passes through the point (0, 0)
    3. C The graph has a vertical asymptote at x = 0
    4. D The graph lies entirely below the x-axis
  3. 31 mark

    Given that log_5(x) = -2, find the value of x.

    1. A -10
    2. B 1/10
    3. C -1/25
    4. D 1/25
  4. 41 mark

    Given that log_2(3) = p, express log_2(24) in terms of p.

    1. A 3p
    2. B p + 3
    3. C p + 8
    4. D p - 3
  5. 51 mark

    Given that log_a(x) = 3, find log_a(x^2 / a).

    1. A 6
    2. B 8
    3. C 5
    4. D -5
  6. 61 mark

    Evaluate log_4(1/8).

    1. A 3/2
    2. B -2
    3. C -2/3
    4. D -3/2
  7. 71 mark

    Solve 8^(x - 2) = 1/16 for x.

    1. A 2/3
    2. B -2/3
    3. C 10/3
    4. D -1
  8. 81 mark

    How many real solutions does the equation 4^x - 6(2^x) + 9 = 0 have?

    1. A 0
    2. B 1
    3. C 2
    4. D infinitely many
  9. 91 mark

    Find x, given that 4^x - 3(2^x) - 4 = 0.

    1. A -1
    2. B 0
    3. C 4
    4. D 2
  10. 101 mark

    Given that log_6(2) = t, express log_6(18) in terms of t.

    1. A 3t
    2. B 1
    3. C 2 - t
    4. D -t
  11. 111 mark

    The graphs of y = 2^x and y = (1/2)^x intersect at exactly one point. Find the y-coordinate of this point.

    1. A 2
    2. B 1
    3. C 0
    4. D 1/2
  12. 121 mark

    Solve log_2(x) + log_2(x - 2) = 3 for x.

    1. A 4
    2. B -2
    3. C 3
    4. D 5

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. Substituting each point into y = a^x gives two equations: a^2 = 9 and a^5 = 243.
    2. Dividing the second equation by the first avoids solving for a from a single point directly: a^5 / a^2 = a^3 = 243/9 = 27.
    3. Since a > 0, taking the cube root gives a unique real value: a = 3 (check: 3^2 = 9, matching the first point, and 3^5 = 243, matching the second).
    4. So a = 3, option C.
    • Why not A: Treats the ratio a^3 = 243/9 as a subtraction instead of a division, computing 243 - 9 = 234 rather than 243/9.
    • Why not B: Divides in the wrong order, computing a^3 = 9/243 = 1/27, and then stops without taking the cube root, giving the value of a^3 rather than a.
    • Why not D: Correctly reaches a^3 = 27 but forgets to take the cube root, treating the cubed value itself as the final answer for a.

  2. Question 2Answer: A

    1. Since 0 < 3/7 < 1, raising 3/7 to larger and larger powers of x produces smaller and smaller positive values, so the graph falls continuously as x increases: it is decreasing for all real x.
    2. At x = 0, y = (3/7)^0 = 1, so the graph passes through (0, 1), ruling out (0, 0).
    3. An exponential function y = a^x is never zero and never negative for a > 0, so the graph stays strictly above the x-axis (with a horizontal asymptote at y = 0), and has no vertical asymptote at all.
    4. So the true statement is that the graph is decreasing for all real x, option A.
    • Why not B: Confuses the y-intercept of an exponential graph with that of a graph through the origin: since (3/7)^0 = 1, the graph passes through (0, 1), not (0, 0).
    • Why not C: Confuses the behaviour of an exponential graph with that of a reciprocal graph such as y = 1/x. An exponential graph has a horizontal asymptote (at y = 0), not a vertical one.
    • Why not D: Confuses a decreasing graph with a negative one. Since 3/7 > 0, every value of (3/7)^x is positive, so the graph lies entirely above the x-axis, however small its values become.

  3. Question 3Answer: D

    1. The statement log_a(b) = c means the same thing as a^c = b, so log_5(x) = -2 means 5^(-2) = x.
    2. A negative exponent means take the reciprocal of the corresponding positive power: 5^(-2) = 1/(5^2) = 1/25.
    3. So x = 1/25, option D. Since x is defined here as a power of a positive base, it must itself be positive, ruling out any negative option.
    • Why not A: Treats the equation log_5(x) = -2 as a linear equation in x, computing 5 x (-2) = -10 instead of recognising that -2 is the exponent, not a coefficient to multiply by.
    • Why not B: Misreads the meaning of a negative exponent, computing 5^(-2) as 1/(5 x 2) = 1/10 instead of 1/(5^2) = 1/25.
    • Why not C: Correctly computes the size of the answer as 1/25 but wrongly carries the negative sign of the exponent through to the value of x itself, forgetting that x = 5^(-2) is a positive number since 5^(-2) means 1/(5^2), not -(5^2).

  4. Question 4Answer: B

    1. Factorise 24 to relate it to a power of 2 and to 3: 24 = 8 x 3 = 2^3 x 3.
    2. Apply the product law: log_2(2^3 x 3) = log_2(2^3) + log_2(3).
    3. log_2(2^3) = 3, since raising the base 2 to the power 3 gives 2^3 directly.
    4. Substituting log_2(3) = p gives log_2(24) = 3 + p, which is the same as p + 3, option B.
    • Why not A: Correctly finds that 24 = 8 x 3 and that log_2(8) = 3, but then combines the two terms by multiplying instead of adding, treating the product law log_a(mn) = log_a(m) + log_a(n) as though it read log_a(mn) = log_a(m) x log_a(n).
    • Why not C: Correctly splits 24 into 8 x 3, but confuses log_2(8) with the number 8 itself, forgetting that log_2(8) = 3, not 8.
    • Why not D: Uses the quotient law instead of the product law, subtracting the two terms as though 24 were a quotient of 8 and 3 rather than their product.

  5. Question 5Answer: C

    1. Apply the quotient law first: log_a(x^2/a) = log_a(x^2) - log_a(a).
    2. Apply the power law to the first term: log_a(x^2) = 2 log_a(x) = 2 x 3 = 6.
    3. Use the special case log_a(a) = 1 for the second term.
    4. So log_a(x^2/a) = 6 - 1 = 5, option C.
    • Why not A: Applies the power law correctly to get log_a(x^2) = 2 x 3 = 6, but forgets to subtract log_a(a) for the division by a, treating log_a(x^2/a) as though it were just log_a(x^2).
    • Why not B: Confuses the power law k log_a(m) = log_a(m^k) with squaring the log value itself, computing (log_a(x))^2 = 3^2 = 9 instead of 2 log_a(x) = 6, and then subtracts 1 to get 8.
    • Why not D: Reverses the order of the quotient law, computing log_a(a) - log_a(x^2) = 1 - 6 = -5 instead of log_a(x^2) - log_a(a).

  6. Question 6Answer: D

    1. Rewrite both 4 and 8 as powers of 2: 4 = 2^2 and 8 = 2^3.
    2. log_4(8) asks: to what power must 4 = 2^2 be raised to give 8 = 2^3? Since (2^2)^k = 2^(2k), we need 2k = 3, so k = 3/2. Check: 4^(3/2) = (4^(1/2))^3 = 2^3 = 8, confirming log_4(8) = 3/2.
    3. By the special case log_a(1/x) = -log_a(x), log_4(1/8) = -log_4(8) = -3/2, option D.
    • Why not A: Correctly identifies that log_4(8) = 3/2, but forgets the negative sign that the special case log_a(1/x) = -log_a(x) introduces for a reciprocal argument.
    • Why not B: Mistakenly believes 4^2 = 8 (rather than 4^2 = 16), and so reports log_4(8) as 2, giving log_4(1/8) = -2.
    • Why not C: Correctly identifies that the exponent involves a fraction with denominator 2, but inverts it, writing 3/2 as 2/3, and applies the negative sign to this incorrect fraction.

  7. Question 7Answer: A

    1. Rewrite both sides with base 2: 8 = 2^3, so 8^(x-2) = 2^(3(x-2)) = 2^(3x-6). Also 1/16 = 2^(-4).
    2. Matching the bases means the exponents are equal: 3x - 6 = -4.
    3. Add 6 to both sides: 3x = 2.
    4. Divide by 3: x = 2/3, option A.
    • Why not B: Distributes the outer exponent incorrectly across the bracket, multiplying only the x term by 3 and leaving the constant term unchanged, solving 3x - 2 = -4 instead of 3(x - 2) = -4, giving x = -2/3.
    • Why not C: Makes a sign error rewriting the right-hand side, treating 1/16 as 2^4 instead of 2^(-4), and solves 3x - 6 = 4 to get x = 10/3.
    • Why not D: Correctly reaches 3x - 6 = -4 and 3x = 2, but then subtracts 3 instead of dividing by it, computing 2 - 3 = -1 instead of 2/3.

  8. Question 8Answer: B

    1. Let y = 2^x. Since 4 = 2^2, 4^x = (2^x)^2 = y^2, so the equation becomes y^2 - 6y + 9 = 0.
    2. This factorises as a perfect square: (y - 3)^2 = 0, so y = 3 is a repeated root, not two distinct roots.
    3. y = 3 is positive, so it is achievable by y = 2^x: solving 2^x = 3 gives exactly one real value of x.
    4. Since the quadratic in y has only one distinct root, and each valid value of y gives exactly one value of x, the original equation has exactly 1 real solution, option B.
    • Why not A: Misapplies the idea of a discriminant, believing that the equation y^2 - 6y + 9 = 0 (with y = 2^x) has no real roots, when in fact its discriminant is 6^2 - 4(9) = 0, meaning it has exactly one repeated real root, not zero.
    • Why not C: Assumes that any quadratic obtained by substitution must have two distinct roots, without checking that y^2 - 6y + 9 factorises as the perfect square (y - 3)^2, which gives a single repeated root.
    • Why not D: Mistakes the repeated root of the perfect square (y - 3)^2 = 0 for an identity true for a whole range of values of y, rather than a single value y = 3 counted once.

  9. Question 9Answer: D

    1. Let y = 2^x. Since 4 = 2^2, 4^x = y^2, so the equation becomes y^2 - 3y - 4 = 0.
    2. Factorise: (y - 4)(y + 1) = 0, so y = 4 or y = -1.
    3. Since y = 2^x and 2^x is always positive for real x, the root y = -1 must be rejected. Only y = 4 is valid.
    4. Solve 2^x = 4 = 2^2: matching the bases gives x = 2, option D.
    • Why not A: Correctly factorises the quadratic in y = 2^x as (y - 4)(y + 1) = 0, but then uses the invalid root y = -1 directly as the value of x, forgetting that y stands for 2^x, not x itself, and that 2^x can never be negative.
    • Why not B: Makes a sign error factorising the quadratic, obtaining (y + 4)(y - 1) = 0 instead of (y - 4)(y + 1) = 0, and accepts the resulting root y = 1, solving 2^x = 1 to get x = 0.
    • Why not C: Correctly identifies y = 4 as the only valid root but then treats the equation 2^x = 4 as though the base and the target must be directly equal, reporting x = 4 instead of solving 2^x = 2^2 to get x = 2.

  10. Question 10Answer: C

    1. Since 6 = 2 x 3, log_6(2) + log_6(3) = log_6(6) = 1. Substituting log_6(2) = t gives log_6(3) = 1 - t.
    2. Write 18 as 2 x 3^2, so that it splits into factors already related to t: log_6(18) = log_6(2) + log_6(3^2).
    3. Apply the power law to the second term: log_6(3^2) = 2 log_6(3) = 2(1 - t).
    4. Combine: log_6(18) = t + 2(1 - t) = t + 2 - 2t = 2 - t, option C.
    • Why not A: Uses log_6(6) = 1 to find that log_6(3) = 1 - t is needed, but wrongly assumes log_6(3) is equal to t as well (confusing it with the given value for log_6(2)), and computes log_6(18) = log_6(2) + 2log_6(3) = t + 2t = 3t.
    • Why not B: Correctly derives log_6(3) = 1 - t from log_6(2) + log_6(3) = log_6(6) = 1, but forgets to apply the power law for the factor of 3^2 inside 18, treating log_6(9) as log_6(3) with no doubling, giving log_6(2) + log_6(3) = t + (1 - t) = 1.
    • Why not D: Confuses the identity log_6(6) = 1 with log_a(1) = 0, wrongly assuming log_6(2) + log_6(3) = 0 rather than 1, and so derives log_6(3) = -t, giving log_6(18) = log_6(2) + 2log_6(3) = t + 2(-t) = -t.

  11. Question 11Answer: B

    1. Since 1/2 = 2^(-1), the equation y = (1/2)^x can be rewritten as y = 2^(-x).
    2. Setting the two expressions equal: 2^x = 2^(-x). Since y = 2^u is a one-to-one function (no two different exponents give the same value), the exponents themselves must be equal: x = -x.
    3. Solving 2x = 0 gives x = 0.
    4. Substituting back into y = 2^x gives y = 2^0 = 1, so the y-coordinate of the intersection point is 1, option B.
    • Why not A: Assumes without checking that the intersection point occurs at x = 1, and evaluates y = 2^1 = 2, rather than solving 2^x = (1/2)^x algebraically to find the correct x-value first.
    • Why not C: Correctly finds that the graphs meet where x = 0, but confuses this x-coordinate with the y-coordinate being asked for, reporting the value of x instead of y.
    • Why not D: Evaluates the wrong function at the wrong point, computing y = (1/2)^1 = 1/2 instead of solving for the actual point where the two graphs are equal.

  12. Question 12Answer: A

    1. Apply the product law for logarithms: log_2(x) + log_2(x - 2) = log_2[x(x - 2)], so the equation becomes log_2[x(x - 2)] = 3.
    2. Convert out of log form: x(x - 2) = 2^3 = 8.
    3. Expand and rearrange into a quadratic: x^2 - 2x - 8 = 0, which factorises as (x - 4)(x + 2) = 0, so x = 4 or x = -2.
    4. Check both roots against the domain of the original equation: log_2(x) requires x > 0 and log_2(x - 2) requires x > 2, so only x = 4 satisfies both; x = -2 must be rejected since neither logarithm is defined there.
    5. So x = 4, option A.
    • Why not B: Correctly forms the quadratic x^2 - 2x - 8 = 0 and finds both roots x = 4 and x = -2, but takes x = -2 as a valid answer without checking it against the domain of the original equation: log_2(x) and log_2(x - 2) are both undefined when x = -2, since a logarithm's argument must be positive.
    • Why not C: Forgets to convert the right-hand side out of log form, treating log_2[x(x - 2)] = 3 as x(x - 2) = 3 instead of x(x - 2) = 2^3 = 8, and solves the resulting quadratic x^2 - 2x - 3 = 0 to get x = 3 (correctly rejecting the negative root -1, but from the wrong quadratic).
    • Why not D: Misapplies the product law for logarithms, treating log_2(x) + log_2(x - 2) = 3 as though it meant log_2(x + (x - 2)) = 3, and so solves x + (x - 2) = 2^3 = 8 to get 2x - 2 = 8, x = 5, instead of first multiplying the two arguments together.

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