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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 1

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 = 4^x - 5 has a horizontal asymptote at y = k. Find k.

    1. A -5
    2. B -4
    3. C 0
    4. D 5
  2. 21 mark

    The graph of y = 3^x is reflected in the y-axis to give the graph of y = a^x, where a > 0. Find a.

    1. A -3
    2. B 1/3
    3. C -1/3
    4. D 3
  3. 31 mark

    The graph of y = a^x, where a > 0, passes through the point (3, 1/8). Find a.

    1. A 1/24
    2. B 2
    3. C 1/2
    4. D -1/2
  4. 41 mark

    The graph of y = 2^x is translated 4 units in the positive x-direction to give the graph of y = f(x). Find f(0).

    1. A -4
    2. B 16
    3. C 1/2
    4. D 1/16
  5. 51 mark

    Given log_a(5) = x and log_a(2) = y, express log_a(50) in terms of x and y.

    1. A x^2 + y
    2. B 2x + y
    3. C 2xy
    4. D x + 2y
  6. 61 mark

    Given log_3(x) = 4, find log_3(9/x).

    1. A 6
    2. B 2
    3. C -2
    4. D -1
  7. 71 mark

    Given log_7(2) = m, express log_7(8) - log_7(4) in terms of m.

    1. A -m
    2. B 5m
    3. C 6m^2
    4. D m
  8. 81 mark

    Simplify log_9(3) + log_9(1/3).

    1. A 0
    2. B 1
    3. C -1
    4. D 1/2
  9. 91 mark

    Solve 3^(2x) = 4, giving your answer in the form x = log_3(k) for a constant k. Find k.

    1. A 4
    2. B 16
    3. C 2
    4. D 8
  10. 101 mark

    How many real values of x satisfy 9^x - 4(3^x) + 3 = 0?

    1. A 1
    2. B 0
    3. C 4
    4. D 2
  11. 111 mark

    Find the sum of all values of x satisfying 9^x - 12(3^x) + 27 = 0.

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

    Solve for x: (1/4)^x = 8.

    1. A 3/2
    2. B -3/2
    3. C -2/3
    4. D -6

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: A

    1. The graph of y = 4^x has a horizontal asymptote at y = 0, since 4^x gets closer and closer to 0 as x becomes very negative but never reaches or crosses it.
    2. Subtracting 5 from the whole function, y = 4^x - 5, shifts every point on the graph down by 5 units, including the asymptote.
    3. So the asymptote y = 0 shifts to y = 0 - 5 = -5, giving k = -5, option A.
    • Why not B: Confuses the vertical shift with the base of the exponential, giving the negative of the base (4) instead of the negative of the shift (5).
    • Why not C: Believes the horizontal asymptote of an exponential graph is always y = 0 regardless of any vertical translation applied to it.
    • Why not D: Makes a sign error on the translation, treating 'subtract 5' as though it moved the graph up by 5 rather than down by 5.

  2. Question 2Answer: B

    1. Reflecting a graph in the y-axis replaces x with -x in its equation, so y = 3^x becomes y = 3^(-x).
    2. Using the law a^(-n) = (1/a)^n, this can be rewritten as y = (3^(-1))^x = (1/3)^x.
    3. So the reflected graph is y = (1/3)^x, meaning a = 1/3, option B.
    • Why not A: Applies the reflection directly to the base a, negating it to -3, forgetting that the specification requires a to stay positive and that a reflection in the y-axis instead changes the sign of the exponent.
    • Why not C: Combines two separate errors: takes the reciprocal of the base (as the correct method requires) but also negates it, confusing a reflection in the y-axis with a reflection through the origin.
    • Why not D: Believes reflecting y = 3^x in the y-axis leaves the equation unchanged, as would happen for an even function, without checking whether y = 3^x actually has that symmetry.

  3. Question 3Answer: C

    1. Substituting the point (3, 1/8) into y = a^x gives a^3 = 1/8.
    2. A cube root is unique for a real number, unlike a square root, so there is exactly one real value of a satisfying this.
    3. Since (1/2)^3 = 1/8, taking the cube root of both sides gives a = 1/2, option C.
    • Why not A: Confuses the cubic equation a^3 = 1/8 with the linear equation 3a = 1/8, dividing 1/8 by 3 instead of taking a cube root.
    • Why not B: Misreads the fraction 1/8 as 8, and finds the cube root of 8 instead of the cube root of 1/8, forgetting the reciprocal entirely.
    • Why not D: Incorrectly believes a cube root has two real values in the way a square root does, offering the negative alternative despite (-1/2)^3 = -1/8, not 1/8.

  4. Question 4Answer: D

    1. Translating y = 2^x by 4 units in the positive x-direction gives y = f(x) = 2^(x-4), since a shift right by 4 replaces x with (x - 4).
    2. To find f(0), substitute x = 0: f(0) = 2^(0-4) = 2^(-4).
    3. 2^(-4) = 1/(2^4) = 1/16, option D.
    • Why not A: Confuses the amount of the horizontal translation with the value of the translated function, reporting the shift distance itself rather than evaluating f at x = 0.
    • Why not B: Forgets the negative sign that a translation in the positive x-direction introduces into the exponent, computing 2^4 instead of 2^(-4).
    • Why not C: Loses track of the shift amount, using an exponent of -1 instead of -4, and computes 2^(-1) instead of 2^(-4).

  5. Question 5Answer: B

    1. Factorise 50 to see how it relates to 5 and 2: 50 = 5^2 x 2.
    2. Apply the laws of logarithms: log_a(5^2 x 2) = log_a(5^2) + log_a(2) = 2 log_a(5) + log_a(2).
    3. Substituting log_a(5) = x and log_a(2) = y gives 2x + y, option B.
    • Why not A: Confuses the power law k log_a(m) = log_a(m^k) with squaring the log value itself, writing (log_a 5)^2 instead of 2 log_a(5).
    • Why not C: Multiplies the two log terms together instead of adding them, treating a sum of logarithms as though it combined multiplicatively.
    • Why not D: Mis-factorises 50 as 5 x 2^2 (which equals 20, not 50) instead of 5^2 x 2, and so doubles the coefficient of y instead of the coefficient of x.

  6. Question 6Answer: C

    1. Use the quotient law: log_3(9/x) = log_3(9) - log_3(x).
    2. log_3(9) = 2, since 3^2 = 9.
    3. Substituting log_3(x) = 4 gives log_3(9/x) = 2 - 4 = -2, option C.
    • Why not A: Adds instead of subtracting, computing log_3(9) + log_3(x) = 2 + 4 = 6, misapplying the product law to what is actually a quotient.
    • Why not B: Subtracts in the wrong order, computing log_3(x) - log_3(9) = 4 - 2 = 2 instead of log_3(9) - log_3(x).
    • Why not D: Misidentifies log_3(9) as 3 instead of 2, confusing it with log_3(27), and then computes 3 - 4 = -1.

  7. Question 7Answer: D

    1. Notice that 8/4 = 2, so by the quotient law, log_7(8) - log_7(4) = log_7(8/4) = log_7(2) = m directly.
    2. This can also be checked the long way: 8 = 2^3, so log_7(8) = 3 log_7(2) = 3m; 4 = 2^2, so log_7(4) = 2 log_7(2) = 2m.
    3. Subtracting gives 3m - 2m = m, confirming the shorter route, option D.
    • Why not A: Reverses the order of subtraction, computing log_7(4) - log_7(8) = 2m - 3m = -m instead of log_7(8) - log_7(4).
    • Why not B: Adds the two exponent coefficients instead of subtracting them, treating the quotient of two logarithms as though it required the product law: 3m + 2m = 5m.
    • Why not C: Multiplies the two exponent-coefficient expressions together instead of subtracting them, confusing the quotient law with a multiplication of the log terms.

  8. Question 8Answer: A

    1. log_9(3) = 1/2, since 9^(1/2) = 3.
    2. By the special case log_a(1/x) = -log_a(x), log_9(1/3) = -log_9(3) = -1/2.
    3. Adding the two: log_9(3) + log_9(1/3) = 1/2 + (-1/2) = 0, option A.
    • Why not B: Forgets the negative sign in the law log_a(1/x) = -log_a(x), treating log_9(1/3) as equal to log_9(3) rather than as its negative, giving 1/2 + 1/2 = 1.
    • Why not C: Applies a negative sign to both terms instead of just the second, incorrectly treating log_9(3) as -1/2 as well as log_9(1/3), giving -1/2 + (-1/2) = -1.
    • Why not D: Correctly finds log_9(3) = 1/2 but treats log_9(1/3) as 0, mistakenly believing the logarithm of a reciprocal is always 0.

  9. Question 9Answer: C

    1. Take log base 3 of both sides: log_3(3^(2x)) = log_3(4), so 2x = log_3(4).
    2. Divide both sides by 2: x = (1/2) log_3(4).
    3. Use the power law in reverse, k log_a(m) = log_a(m^k), with k = 1/2: (1/2) log_3(4) = log_3(4^(1/2)) = log_3(2), since 4^(1/2) = 2.
    4. So x = log_3(2), meaning k = 2, option C.
    • Why not A: Forgets to divide the exponent equation 2x = log_3(4) by 2, reporting x = log_3(4) directly instead of x = (1/2) log_3(4), giving k = 4.
    • Why not B: Attempts to move the coefficient 1/2 inside the logarithm but squares the argument instead of taking its square root, computing 4^2 = 16 instead of 4^(1/2) = 2.
    • Why not D: Confuses 'divide the outer exponent by 2' with 'multiply the argument by 2', computing 4 x 2 = 8 instead of taking a square root.

  10. Question 10Answer: D

    1. Let y = 3^x. Since 9 = 3^2, we have 9^x = (3^2)^x = (3^x)^2 = y^2, so the equation becomes y^2 - 4y + 3 = 0.
    2. Factorise: (y - 1)(y - 3) = 0, so y = 1 or y = 3.
    3. Both values are positive, and 3^x is always positive, so both are achievable: 3^x = 1 gives x = 0, and 3^x = 3 gives x = 1.
    4. These are two distinct real values of x, option D.
    • Why not A: Mistakenly believes 3^x = 1 has no solution, forgetting that a^0 = 1 for any positive a, and so only counts the solution coming from 3^x = 3.
    • Why not B: Miscalculates the quadratic in y (for example misreading the constant term) and wrongly concludes it has no real roots at all.
    • Why not C: Conflates the two roots of the quadratic in y (1 and 3) with the two values of x they produce (0 and 1), treating all four numbers as separate solutions instead of realising each y-root gives exactly one x-value.

  11. Question 11Answer: A

    1. Let y = 3^x. Since 9 = 3^2, we have 9^x = (3^x)^2 = y^2, so the equation becomes y^2 - 12y + 27 = 0.
    2. Factorise: (y - 3)(y - 9) = 0, so y = 3 or y = 9, both of which are positive and so both achievable by 3^x.
    3. y = 3 gives 3^x = 3, so x = 1. y = 9 gives 3^x = 3^2, so x = 2.
    4. The sum of the two solutions is 1 + 2 = 3, option A.
    • Why not B: Reports the sum of the y-roots of the quadratic (3 + 9 = 12) directly, forgetting to convert each one back to a value of x first.
    • Why not C: Correctly solves 3^x = 3 to get x = 1, but mistakenly believes 3^x = 9 has no solution, failing to recognise that 9 = 3^2, and so omits the second root entirely.
    • Why not D: Solves 3^x = 9 by dividing 9 by 3 (giving x = 9/3 = 3) instead of recognising 9 = 3^2 (giving x = 2), then adds this incorrect value to the correctly found x = 1, giving 1 + 3 = 4.

  12. Question 12Answer: B

    1. Rewrite 1/4 as a power of 2: 1/4 = 2^(-2), so (1/4)^x = (2^(-2))^x = 2^(-2x).
    2. Rewrite 8 as a power of 2: 8 = 2^3, so the equation becomes 2^(-2x) = 2^3.
    3. Matching bases means the exponents must be equal: -2x = 3.
    4. Divide both sides by -2: x = 3/(-2) = -3/2, option B.
    • Why not A: Loses the negative sign when rewriting 1/4 as a power of 2, treating it as 2^2 instead of 2^(-2), and so solves 2x = 3 to get x = 3/2.
    • Why not C: Inverts the final division, computing 3/(-2) as -2/3 instead of -3/2, swapping the numerator and denominator.
    • Why not D: Confuses solving the linear equation -2x = 3 with multiplying instead of dividing, computing -2 x 3 = -6 instead of 3/(-2).

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