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Admissions tests / ESAT / Maths 2 / Exponentials and logarithms

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 2: Exponentials and logarithms, set 1

Exponential graphs, the laws of logarithms, solving exponential equations, 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 = 3^x crosses the y-axis at the point (0, k). Find k.

    1. A 0
    2. B 1
    3. C 3
    4. D 9
  2. 21 mark

    For every value of a > 0, the graph of y = a^x passes through which of these points, regardless of the value of a?

    1. A (1, a)
    2. B (1, 1)
    3. C (0, a)
    4. D (0, 1)
  3. 31 mark

    For a > 1, as x becomes a very large negative number, what happens to y = a^x?

    1. A y gets closer and closer to 0 but stays positive
    2. B y becomes a very large positive number
    3. C y becomes a very large negative number
    4. D y gets closer and closer to 1
  4. 41 mark

    The function y = a^x is decreasing for all real x. Which of these could be the value of a?

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

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

    1. A 4.5
    2. B 3
    3. C 9
    4. D 81
  6. 61 mark

    Given log_a(2) = p and log_a(3) = q, express log_a(6) in terms of p and q.

    1. A p + q
    2. B p - q
    3. C pq
    4. D p/q
  7. 71 mark

    Given log_a(20) = m and log_a(4) = n, express log_a(5) in terms of m and n.

    1. A m + n
    2. B n - m
    3. C mn
    4. D m - n
  8. 81 mark

    Given log_a(3) = t, express log_a(81) in terms of t.

    1. A t^4
    2. B t + 4
    3. C 4t
    4. D 3t
  9. 91 mark

    Given log_2(x) = 5, find log_2(1/x).

    1. A 5
    2. B -5
    3. C 1/5
    4. D -1/5
  10. 101 mark

    Simplify log_6(6) + log_6(36).

    1. A 3
    2. B 2
    3. C 1
    4. D 216
  11. 111 mark

    Solve for x: 5^x = 125.

    1. A 4
    2. B 25
    3. C 5
    4. D 3
  12. 121 mark

    Solve for x: 5^(2x) = 125.

    1. A 3
    2. B 6
    3. C 3/2
    4. D 1
  13. 131 mark

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

    1. A -3
    2. B 3
    3. C 2
    4. D 1
  14. 141 mark

    Find the sum of all solutions x to the equation 2^(2x) - 5(2^x) + 4 = 0.

    1. A 2
    2. B 5
    3. C 4
    4. D 3
  15. 151 mark

    Solve for x: 2^(3x) = 4^(x + 1).

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

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: B

    1. The y-intercept occurs where x = 0.
    2. Any positive number raised to the power 0 equals 1, so 3^0 = 1.
    3. Therefore the graph crosses the y-axis at (0, 1), so k = 1, option B.
    • Why not A: Believes any number raised to the power 0 equals 0, evaluating 3^0 = 0 instead of 1.
    • Why not C: Evaluates 3^1 instead of 3^0, mistaking the y-intercept (where x = 0) for the value of the function at x = 1.
    • Why not D: Evaluates 3^2 instead of 3^0, misreading the y-intercept as if it required squaring the base.

  2. Question 2Answer: D

    1. For any positive base a, a^0 = 1, so every graph y = a^x passes through (0, 1) regardless of what a is.
    2. The point (1, a) does lie on the graph, but its y-coordinate is different for each value of a, so it is not a fixed point.
    3. The point that is the same for every a is therefore (0, 1), option D.
    • Why not A: Uses the true fact that a^1 = a to name the point (1, a), but this point's y-coordinate changes with a, so it is not the same point for every value of a.
    • Why not B: Confuses the rule a^0 = 1 with a different rule, incorrectly believing a^1 = 1 for every base a.
    • Why not C: Believes a^0 = a rather than a^0 = 1, misapplying the base value to the y-intercept.

  3. Question 3Answer: A

    1. For a > 1, increasing x makes a^x grow; decreasing x (making it very negative) makes a^x shrink towards 0.
    2. Since a^x = 1 / a^(-x), a very negative x makes -x very large and positive, so a^x becomes 1 divided by a very large number, which is a small positive number.
    3. a^x can never reach or go below 0, since a power of a positive number is always positive, so y approaches 0 but stays positive, option A.
    • Why not B: Confuses the behaviour as x tends to negative infinity with the behaviour as x tends to positive infinity, where y does grow without bound; forgets that a very negative exponent instead makes y small.
    • Why not C: Assumes that a very negative exponent must produce a very negative output, but a^x is positive for every real x regardless of the sign of x.
    • Why not D: Confuses the horizontal asymptote (y = 0) with the y-intercept value (y = 1 at x = 0).

  4. Question 4Answer: C

    1. For y = a^x with 0 < a < 1, increasing x makes the value shrink towards 0, so the function is decreasing.
    2. For a > 1 the function increases, and for a = 1 it is constant, so neither gives a decreasing function.
    3. Only a = 1/3, which satisfies 0 < a < 1, gives a decreasing exponential function, option C.
    • Why not A: Picks a value of a greater than 1, which makes y = a^x an increasing function, not a decreasing one.
    • Why not B: Picks a = 1, which makes y = 1^x = 1 for every x, a constant function rather than a decreasing one.
    • Why not D: Picks a negative value, overlooking that the specification restricts a to positive values; a^x is not a smoothly increasing or decreasing curve when a is negative.

  5. Question 5Answer: B

    1. Substituting the point (2, 9) into y = a^x gives a^2 = 9.
    2. Since a > 0, take the positive square root of both sides: a = sqrt(9) = 3.
    3. So a = 3, option B.
    • Why not A: Treats the equation a^2 = 9 as if it were 2a = 9, dividing 9 by 2 to get 4.5 instead of taking a square root.
    • Why not C: Reads off a = 9 directly from the point, forgetting that the equation to solve is a^2 = 9, not a = 9.
    • Why not D: Squares 9 instead of taking its square root, computing 9^2 = 81 instead of undoing the square on a^2 = 9.

  6. Question 6Answer: A

    1. 6 = 2 x 3, so use the product law: log_a(mn) = log_a(m) + log_a(n).
    2. log_a(6) = log_a(2 x 3) = log_a(2) + log_a(3) = p + q.
    3. So log_a(6) = p + q, option A.
    • Why not B: Applies the quotient law instead of the product law, subtracting the two logs instead of adding them.
    • Why not C: Multiplies the two log values together, confusing the multiplication of the numbers inside the log (2 x 3) with a multiplication of the log values themselves.
    • Why not D: Divides the two log values, confusing division of the numbers inside the log with a division of the log values themselves.

  7. Question 7Answer: D

    1. 5 = 20/4, so use the quotient law: log_a(m/n) = log_a(m) - log_a(n).
    2. log_a(5) = log_a(20/4) = log_a(20) - log_a(4) = m - n.
    3. So log_a(5) = m - n, option D.
    • Why not A: Applies the product law instead of the quotient law, adding the two logs instead of subtracting them.
    • Why not B: Subtracts in the wrong order, computing log_a(4) - log_a(20) instead of log_a(20) - log_a(4).
    • Why not C: Multiplies the two log values together, confusing division of the numbers inside the log (20/4) with a multiplication of the log values themselves.

  8. Question 8Answer: C

    1. 81 = 3^4, so log_a(81) = log_a(3^4).
    2. Use the power law: log_a(m^k) = k log_a(m), so log_a(3^4) = 4 log_a(3) = 4t.
    3. So log_a(81) = 4t, option C.
    • Why not A: Raises t to the 4th power instead of multiplying it by 4, confusing the power law log_a(m^k) = k log_a(m) with an exponent applied to the log value itself.
    • Why not B: Adds 4 to t rather than multiplying by it, treating the power law as if the exponent were simply added on to the log value.
    • Why not D: Multiplies t by the base number 3 instead of by the exponent 4, confusing which number in 3^4 = 81 is the power to bring down.

  9. Question 9Answer: B

    1. Use the law log_a(1/x) = -log_a(x), which follows from writing 1/x as x^(-1) and applying the power law.
    2. Since log_2(x) = 5, substituting gives log_2(1/x) = -log_2(x) = -5.
    3. So log_2(1/x) = -5, option B.
    • Why not A: Forgets the negative sign in the law log_a(1/x) = -log_a(x), treating log_a(1/x) as equal to log_a(x).
    • Why not C: Takes the reciprocal of the log value instead of its negative, confusing the reciprocal of x with the reciprocal of log(x).
    • Why not D: Combines both errors above, taking the negative reciprocal of the log value instead of simply its negative.

  10. Question 10Answer: A

    1. log_6(6) = 1, using the rule log_a(a) = 1.
    2. 36 = 6^2, so log_6(36) = 2.
    3. Add the two results: log_6(6) + log_6(36) = 1 + 2 = 3, option A.
    • Why not B: Correctly finds log_6(36) = 2 but treats log_6(6) as 0, mixing up the rule log_a(a) = 1 with the separate rule log_a(1) = 0.
    • Why not C: Correctly finds log_6(6) = 1 but treats log_6(36) as 0, as if only an argument exactly equal to the base can contribute a nonzero value, and so drops that term entirely.
    • Why not D: Multiplies the two arguments (6 x 36 = 216) instead of adding their logarithms, as if the question asked for 6 x 36 rather than log_6(6) + log_6(36).

  11. Question 11Answer: D

    1. 5^x = 125 asks: to what power must 5 be raised to give 125?
    2. 5^1 = 5, 5^2 = 25, 5^3 = 125.
    3. So 125 = 5^3, meaning x = 3, option D.
    • Why not A: Lists 1, 5, 25, 125 as the first four powers of 5 (starting from 5^0 = 1) and mistakenly counts 125 as the 4th power, one too many, instead of the 3rd.
    • Why not B: Treats solving a^x = b as dividing b by a, computing 125/5 = 25 and giving this as the value of x instead of finding the required exponent.
    • Why not C: Confuses the exponent being solved for with the base of the equation, giving x = 5 (the base) instead of the power to which 5 must be raised.

  12. Question 12Answer: C

    1. Write 125 as a power of 5: 125 = 5 x 5 x 5 = 5^3.
    2. So 5^(2x) = 5^3 has matching bases, which means the exponents must be equal: 2x = 3.
    3. Divide both sides by 2 to isolate x: x = 3/2, option C.
    • Why not A: Sets x equal to the exponent 3 directly, forgetting to divide by 2, so solves 2x = 3 as if it were x = 3.
    • Why not B: Multiplies instead of dividing when isolating x, computing 2x = 3 as x = 3 x 2 = 6.
    • Why not D: Confuses 125 with 5^2, miscounting the powers of 5 (5, 25) and stopping one power short, so treats 125 as the 2nd power instead of the 3rd and solves 2x = 2 to get x = 1.

  13. Question 13Answer: B

    1. Rewrite both sides with base 2: 4 = 2^2 and 8 = 2^3, so 4^x = 2^(2x) and 8^(x - 1) = 2^(3(x - 1)) = 2^(3x - 3).
    2. Since the bases now match, the exponents must be equal: 2x = 3x - 3.
    3. Rearrange: 2x - 3x = -3, so -x = -3, giving x = 3, option B.
    • Why not A: Makes a sign error rearranging the exponent equation, writing 2x - 3x = 3 instead of 2x - 3x = -3, and so solves -x = 3 to get x = -3.
    • Why not C: Misreads the exponential notation as multiplication, treating 4^x = 8^(x - 1) as the linear equation 4x = 8(x - 1) instead of rewriting both sides with base 2, and solves 4x = 8x - 8 to get x = 2.
    • Why not D: Distributes the exponent 3 only over the x-term when rewriting 8^(x - 1) as 2^(3x - 1), instead of correctly multiplying the whole bracket to get 2^(3x - 3), and so solves 2x = 3x - 1 to get x = 1.

  14. Question 14Answer: A

    1. Let y = 2^x. The equation becomes y^2 - 5y + 4 = 0.
    2. Factorise: (y - 1)(y - 4) = 0, so y = 1 or y = 4.
    3. Convert back to x: 2^x = 1 gives x = 0 (since a^0 = 1 for any positive a); 2^x = 4 = 2^2 gives x = 2.
    4. The sum of all solutions is 0 + 2 = 2, option A.
    • Why not B: Adds the two roots of the quadratic in y (1 + 4 = 5) instead of converting each back to x first, forgetting that y = 2^x still needs to be solved for x.
    • Why not C: Makes an error factorising y^2 - 5y + 4 = 0, treating it as a repeated root (y - 4)^2 = 0 and so finding only x = 2, counted twice, giving a sum of 4 instead of including the x = 0 solution.
    • Why not D: Solves 2^x = 1 as x = 1, forgetting that a^0 = 1 for any positive a and that the true solution is x = 0; adding this incorrect x = 1 to the correctly found x = 2 gives 3 instead of 2.

  15. Question 15Answer: C

    1. Rewrite 4 as a power of 2: 4 = 2^2, so 4^(x + 1) = (2^2)^(x + 1) = 2^(2(x + 1)) = 2^(2x + 2).
    2. The equation 2^(3x) = 2^(2x + 2) now has matching bases, so the exponents are equal: 3x = 2x + 2.
    3. Subtract 2x from both sides: x = 2, option C.
    • Why not A: Distributes the exponent only over the x-term when rewriting 4^(x + 1) as 2^(2x + 1) instead of the correct 2^(2x + 2), and so solves 3x = 2x + 1 to get x = 1.
    • Why not B: Moves the 2x term across the equals sign without changing its sign, writing 3x + 2x = 2 instead of 3x - 2x = 2, and solves 5x = 2 to get x = 2/5.
    • Why not D: Sets the exponents equal without first rewriting 4 as a power of 2, treating the equation as if both sides already shared a base and solving 3x = x + 1 directly, giving x = 1/2.

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