Revision Libraryv1.6.8
Choose a subject →
Search
WorksheetsBooksFreePracticeHow it worksPricingFreeAboutFAQ
Choose a subject →

Admissions tests / ESAT / Maths 2 / Exponentials and logarithms

Demanding. 15 questions, 15 marks, about 27 minutes.

ESAT Mathematics 2: Exponentials and logarithms, set 2

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.
  • This is a demanding set: several questions combine more than one idea and need multi-step working, with less signposting than the real exam.
  1. 11 mark

    The graph of y = a^x, where a > 0, passes through the point (-2, 1/9). Find the value of a.

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

    Given that log_a(x) = 3, find log_a(1/x^2).

    1. A 6
    2. B -6
    3. C -3/2
    4. D 1/6
  3. 31 mark

    The graphs of y = 2^x and y = 16 - 2^x are sketched for x >= 0. Find the x-coordinate of the point where they intersect.

    1. A 4
    2. B 2
    3. C 3
    4. D There is no real solution
  4. 41 mark

    Which of the following is true for all a > 0 (a != 1) and all x, y > 0?

    1. A log_a(xy) = log_a(x) - log_a(y)
    2. B log_a(x + y) = log_a(x) + log_a(y)
    3. C log_a(xy) = log_a(x) + log_a(y)
    4. D log_a(xy) = log_a(x) x log_a(y)
  5. 51 mark

    Which of the following statements about the graph of y = 7^x is true?

    1. A It crosses the y-axis at the point (0, 7)
    2. B It has a horizontal asymptote at y = 0
    3. C It has a vertical asymptote at x = 0
    4. D It is a decreasing function for all real x
  6. 61 mark

    Solve the equation 5^(x + 1) - 5^x = 100 for x.

    1. A There is no real solution
    2. B 25
    3. C 5
    4. D 2
  7. 71 mark

    Given that log_a(5) = t, express log_a(sqrt(a)/5) in terms of t.

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

    Solve the equation 4^x = 1/(2^(x - 6)) for x.

    1. A 2
    2. B -6
    3. C 3
    4. D -2
  9. 91 mark

    Solve the equation log_3(x) + log_3(x - 2) = 1 for x, given that x is real.

    1. A 3
    2. B -1
    3. C 1
    4. D 2.5
  10. 101 mark

    For a > b > 1, the curves y = a^x and y = b^x are sketched on the same axes. Which of the following is true for all x > 0?

    1. A a^x < b^x
    2. B a^x = b^x
    3. C a^x > b^x
    4. D a^x > b^x only when x is a whole number
  11. 111 mark

    Solve the equation 3^(2x) - 3^x - 6 = 0 for x, given that x is real.

    1. A x = 1 or x = -2
    2. B x = 1 only
    3. C x = -1 only
    4. D No real solutions
  12. 121 mark

    Which of the following is a correct application of the laws of logarithms, given a > 0, a != 1, and x > 0?

    1. A log_a(1/x) = 1/log_a(x)
    2. B log_a(a) = 0
    3. C log_a(1) = 1
    4. D log_a(1/x) = -log_a(x)
  13. 131 mark

    The equation 2^x + 8 x 2^(-x) = 6 has two real solutions for x. Find the sum of these solutions.

    1. A 3
    2. B 6
    3. C There are no real solutions
    4. D 2
  14. 141 mark

    The graph of y = 4^x is reflected in the x-axis to give the graph of y = g(x). Given that g(x) = -64, find the value of x.

    1. A 16
    2. B 3
    3. C 4
    4. D There is no real solution
  15. 151 mark

    Solve the equation sqrt(2^x) = 16 for x.

    1. A 4
    2. B 2
    3. C 8
    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: D

    1. Substitute the point into y = a^x: a^(-2) = 1/9.
    2. Rewrite the negative exponent as a reciprocal: 1/a^2 = 1/9, so a^2 = 9.
    3. Since a > 0, take the positive square root: a = 3, option D.
    • Why not A: Sets up the equation correctly as a^2 = 9 but then, instead of taking the square root, takes the reciprocal (treating the negative exponent as a sign that the final answer must also be inverted), giving 1/3 instead of 3.
    • Why not B: Finds a^2 = 9 correctly but takes the negative square root, overlooking that a must be positive for y = a^x to be a valid exponential graph.
    • Why not C: Correctly inverts a^(-2) = 1/9 to get a^2 = 9 but stops there without taking the square root, giving the squared value instead of a itself.

  2. Question 2Answer: B

    1. 1/x^2 = x^(-2), so log_a(1/x^2) = log_a(x^(-2)).
    2. By the power law, log_a(x^(-2)) = -2 log_a(x).
    3. Since log_a(x) = 3, this gives -2 x 3 = -6, option B.
    • Why not A: Applies the power law correctly to get 2 log_a(x) = 6 but forgets the negative sign that comes from writing 1/x^2 as x^(-2) rather than x^2.
    • Why not C: Treats the square in the denominator as if it required dividing log_a(x) by 2, as though 1/x^2 meant 1/sqrt(x), giving -log_a(x)/2 = -3/2 instead of multiplying by 2.
    • Why not D: Believes log_a(1/x^2) means taking the reciprocal of log_a(x^2), giving 1/(2 x 3) = 1/6, confusing the reciprocal of the argument with a reciprocal applied to the log value.

  3. Question 3Answer: C

    1. At the point of intersection, 2^x = 16 - 2^x.
    2. Add 2^x to both sides: 2(2^x) = 16, so 2^x = 8.
    3. Since 8 = 2^3, x = 3, option C.
    • Why not A: Sets 2^x equal to 16 directly, forgetting that the right-hand side is 16 minus 2^x rather than 16 alone, and solves 2^x = 16 to get x = 4.
    • Why not B: Treats doubling 2^x as squaring its exponent, rewriting 2(2^x) as 2^(2x) instead of 2 x 2^x, and solves 2^(2x) = 16 = 2^4 to get 2x = 4, so x = 2.
    • Why not D: Moves the 2^x term across the equals sign with the wrong sign, writing 2^x - 2^x = 16 instead of 2^x + 2^x = 16, which simplifies to the false statement 0 = 16 and is wrongly taken to mean the graphs never meet.

  4. Question 4Answer: C

    1. The product law states log_a(mn) = log_a(m) + log_a(n) for any positive m, n and any valid base a.
    2. Applying this with m = x and n = y gives log_a(xy) = log_a(x) + log_a(y).
    3. This holds for every valid x, y and base a, so option C is the true statement.
    • Why not A: Confuses the product law with the quotient law, subtracting the two logarithms instead of adding them.
    • Why not B: Wrongly assumes the product law applies to a sum inside the logarithm, adding x and y before taking the log rather than multiplying them.
    • Why not D: Confuses multiplying the two arguments x and y with multiplying the two logarithm values themselves.

  5. Question 5Answer: B

    1. For y = a^x with a > 0, as x becomes very negative, a^x gets closer and closer to 0 but never reaches or crosses it, so the graph has a horizontal asymptote at y = 0.
    2. The graph has no vertical asymptote, since a^x is defined and finite for every real value of x.
    3. Since 7 > 1, y = 7^x increases as x increases, so it is not a decreasing function.
    4. The only correct statement is therefore that it has a horizontal asymptote at y = 0, option B.
    • Why not A: Confuses the y-intercept with the value of the function at x = 1, evaluating 7^0 as 7 instead of the correct 7^0 = 1.
    • Why not C: Confuses this graph with y = log_7(x), which does have a vertical asymptote at x = 0; y = 7^x is defined and finite for every real x, so it has no vertical asymptote.
    • Why not D: Confuses this graph with the case 0 < a < 1; since 7 > 1, the function y = 7^x is increasing for all real x, not decreasing.

  6. Question 6Answer: D

    1. Write 5^(x+1) as 5 x 5^x, using the index law a^(m+1) = a x a^m.
    2. Substitute: 5 x 5^x - 5^x = 4 x 5^x = 100, so 5^x = 25.
    3. Since 25 = 5^2, x = 2, option D.
    • Why not A: Incorrectly expands 5^(x + 1) as 5^x + 5 (treating addition in the exponent as addition of terms), so the 5^x terms cancel to give the false statement 5 = 100, and wrongly concludes the equation has no solution.
    • Why not B: Correctly reduces the equation to 5^x = 25 but then states x = 25 directly, forgetting that x is the exponent needed to produce 25, not the value 25 itself.
    • Why not C: Correctly reduces the equation to 5^x = 25 but then divides 25 by the base, computing x = 25/5 = 5, instead of recognising that 25 = 5^2.

  7. Question 7Answer: B

    1. Split the quotient using the quotient law: log_a(sqrt(a)/5) = log_a(sqrt(a)) - log_a(5).
    2. Write sqrt(a) as a^(1/2) and use the power law: log_a(a^(1/2)) = (1/2) log_a(a).
    3. Since log_a(a) = 1, this gives (1/2)(1) = 1/2.
    4. So log_a(sqrt(a)/5) = 1/2 - log_a(5) = 1/2 - t, option B.
    • Why not A: Uses log_a(a) = 1 for the numerator but forgets that sqrt(a) = a^(1/2), omitting the power-law factor of 1/2 and treating log_a(sqrt(a)) as log_a(a) = 1 instead of 1/2.
    • Why not C: Applies the quotient law with the two terms in the wrong order, computing log_a(5) - log_a(sqrt(a)) instead of log_a(sqrt(a)) - log_a(5).
    • Why not D: Uses the product law instead of the quotient law, adding log_a(5) rather than subtracting it.

  8. Question 8Answer: A

    1. Write both sides with base 2: 4^x = (2^2)^x = 2^(2x).
    2. Taking the reciprocal flips the sign of the exponent: 1/2^(x - 6) = 2^(-(x - 6)) = 2^(6 - x).
    3. Since the bases now match, the exponents are equal: 2x = 6 - x.
    4. Add x to both sides: 3x = 6, so x = 2, option A.
    • Why not B: Forgets to negate the exponent when taking the reciprocal, treating 1/2^(x - 6) as 2^(x - 6) directly instead of 2^(6 - x), and solves 2x = x - 6 to get x = -6.
    • Why not C: Converts 4^x incorrectly as 2^x, forgetting that 4 = 2^2, instead of the correct 2^(2x), and solves x = 6 - x to get x = 3.
    • Why not D: Negates only the x-term when taking the reciprocal, writing 1/2^(x - 6) as 2^(-x - 6) instead of 2^(6 - x), and solves 2x = -x - 6 to get x = -2.

  9. Question 9Answer: A

    1. Combine the two logarithms using the product law: log_3(x) + log_3(x - 2) = log_3(x(x - 2)) = 1.
    2. Convert to exponential form: x(x - 2) = 3^1 = 3, so x^2 - 2x - 3 = 0.
    3. Factorise: (x - 3)(x + 1) = 0, giving x = 3 or x = -1.
    4. Both log_3(x) and log_3(x - 2) require x > 2, so x = -1 is rejected and only x = 3 is valid, option A.
    • Why not B: Solves the quadratic x^2 - 2x - 3 = 0 correctly to get x = 3 or x = -1, but reports the negative root without checking that log_3(x) and log_3(x - 2) both require x > 2.
    • Why not C: Makes a sign error factorising x^2 - 2x - 3, writing it as (x - 1)(x + 3) instead of (x - 3)(x + 1), and selects the positive root x = 1 without checking it against the domain requirement x > 2.
    • Why not D: Misapplies the product law as if the arguments should be added rather than multiplied, solving x + (x - 2) = 3 instead of x(x - 2) = 3, giving 2x - 2 = 3 and x = 2.5.

  10. Question 10Answer: C

    1. For any x > 0, raising a larger positive base to a positive power gives a larger result: if a > b > 1, then a^x > b^x for every x > 0.
    2. The two curves meet only at x = 0, since a^0 = b^0 = 1; for x > 0 they separate, with the curve of the larger base lying above the other.
    3. This holds for every positive real x, not only whole numbers, so option C is correct.
    • Why not A: Assumes a larger base always gives a smaller value once x is positive, as happens when a base is compared with its reciprocal (for example 2^x against (1/2)^x), without checking that here both a and b are greater than 1, so no such reversal occurs.
    • Why not B: Assumes the two curves coincide for x > 0 because they meet at x = 0, where both equal 1, overlooking that they separate for every x != 0.
    • Why not D: Unnecessarily restricts a general inequality to whole-number values of x, as though raising a larger base to a fractional power did not preserve the same ordering.

  11. Question 11Answer: B

    1. Let y = 3^x. The equation becomes y^2 - y - 6 = 0.
    2. Factorise: (y - 3)(y + 2) = 0, so y = 3 or y = -2.
    3. Since 3^x > 0 for every real x, the value y = -2 is impossible and must be rejected; only y = 3 gives a valid equation.
    4. Solving 3^x = 3 gives x = 1, option B.
    • Why not A: Solves the quadratic in y correctly to get y = 3 or y = -2, correctly converts y = 3 back to x = 1, but for the second root treats y = -2 as if it were itself the value of x, instead of recognising that 3^x = -2 has no real solution and must be rejected, giving the spurious extra solution x = -2 alongside the genuine x = 1.
    • Why not C: Correctly reduces to 3^x = 3 but then makes a sign error, writing x = -1 instead of x = 1, as though 3^(-1) rather than 3^1 equalled 3.
    • Why not D: Makes a sign error in the discriminant, computing b^2 - 4ac as (-1)^2 - 4(1)(6) = -23 by using c = 6 instead of the correct c = -6, wrongly concluding the quadratic in y has no real roots.

  12. Question 12Answer: D

    1. Writing 1/x as x^(-1) and applying the power law gives log_a(1/x) = log_a(x^(-1)) = -log_a(x).
    2. The two special cases are log_a(a) = 1, since a must be raised to the power 1 to give a, and log_a(1) = 0, since a must be raised to the power 0 to give 1; these are opposite results and easy to swap.
    3. So the correct statement among the four is log_a(1/x) = -log_a(x), option D.
    • Why not A: Confuses the reciprocal of x with a reciprocal applied to the logarithm value itself, rather than using log_a(1/x) = -log_a(x).
    • Why not B: Swaps the two special-case results, giving log_a(a) the value that actually belongs to log_a(1); log_a(a) = 1, not 0.
    • Why not C: Swaps the two special-case results the other way round, giving log_a(1) the value that actually belongs to log_a(a); log_a(1) = 0, not 1.

  13. Question 13Answer: A

    1. Let y = 2^x, so 2^(-x) = 1/y. The equation becomes y + 8/y = 6.
    2. Multiply every term by y: y^2 + 8 = 6y, so y^2 - 6y + 8 = 0.
    3. Factorise: (y - 2)(y - 4) = 0, giving y = 2 or y = 4, both of which are valid values of 2^x.
    4. Solving 2^x = 2 gives x = 1; solving 2^x = 4 gives x = 2. The sum of all solutions is 1 + 2 = 3, option A.
    • Why not B: Correctly solves the quadratic in y to get y = 2 and y = 4, but reports the sum of these y-values (2 + 4 = 6) instead of converting each back to x first and summing the x-values (1 + 2 = 3).
    • Why not C: Clears the fraction incorrectly, rewriting 8/y as 8y instead of multiplying every term by y, which turns the equation into y^2 + 8y = 6y, reducing to y(y + 2) = 0 and giving y = 0 or y = -2; since neither is a valid value of 2^x, this is wrongly taken to mean the original equation has no solution.
    • Why not D: Correctly finds x = 1 and x = 2 but reports their product (1 x 2 = 2) instead of their sum, confusing the two.

  14. Question 14Answer: B

    1. Reflecting y = 4^x in the x-axis negates the whole function: g(x) = -4^x.
    2. Setting g(x) = -64 gives -4^x = -64, so 4^x = 64.
    3. Since 64 = 4^3, x = 3, option B.
    • Why not A: Correctly reaches the equation 4^x = 64 but misreads the exponential notation as multiplication, solving 4x = 64 to get x = 16 instead of finding the power to which 4 must be raised.
    • Why not C: Miscounts the powers of 4 (4, 16, 64), treating 64 as the 4th power of 4 instead of the 3rd, and so gives x = 4 instead of x = 3.
    • Why not D: Reflects the graph in the y-axis instead of the x-axis, replacing x with -x rather than negating the function, giving g(x) = 4^(-x); since 4^(-x) is always positive, setting it equal to -64 wrongly appears to have no solution.

  15. Question 15Answer: C

    1. Write the square root as a power: sqrt(2^x) = (2^x)^(1/2) = 2^(x/2).
    2. So the equation becomes 2^(x/2) = 16.
    3. Since 16 = 2^4, x/2 = 4.
    4. Multiply both sides by 2: x = 8, option C.
    • Why not A: Ignores the square root entirely, treating sqrt(2^x) as 2^x, and solves 2^x = 16 = 2^4 to get x = 4 instead of accounting for the power of 1/2 from the square root.
    • Why not B: Confuses taking a square root with squaring, doubling the exponent instead of halving it: sets 2^(2x) = 16 = 2^4, giving 2x = 4, so x = 2.
    • Why not D: Miscounts the powers of 2 (2, 4, 8, 16), treating 16 as the 3rd power of 2 instead of the 4th, and so solves x/2 = 3 to get x = 6.

More on this strand

More free ESAT practice

Every strand of the published ESAT specification, with worked solutions throughout.

Choose a subject →