Hard A-Level Logarithms Questions

Challenging, exam-style A-Level Logarithms questions with worked solutions. Stretch yourself on the hardest hidden quadratic, exponentials, substitution, natural log problems.

hidden quadraticexponentialssubstitutionnatural logproduct lawsolving log equations
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Solve the simultaneous equations log⁡2x+log⁡2y=5\log_2 x + \log_2 y = 5 and (log⁡2x)(log⁡2y)=6(\log_2 x)(\log_2 y) = 6.
Show worked solution

Worked solution

  1. Introduce two letters

    Let a=log⁡2x,  b=log⁡2y\text{Let } a = \log_2 x,\; b = \log_2 y

    Working with a and b turns the log equations into simple number equations.

  2. Translate the first equation

    a+b=5a + b = 5

    log_2 x + log_2 y is a + b, which equals 5.

  3. Translate the second equation

    ab=6ab = 6

    The product of the two logs is ab, which equals 6.

  4. Recognise sum and product of roots

    t2−(a+b)t+ab=0t^2 - (a+b)t + ab = 0

    If we know the sum and product of two numbers, they are the roots of this quadratic. This idea comes from quadratics.

  5. Form the quadratic

    t2−5t+6=0t^2 - 5t + 6 = 0

    Substitute the sum 5 and product 6.

  6. Factorise

    (t−2)(t−3)=0(t-2)(t-3) = 0

    Two numbers multiply to 6 and add to 5: they are 2 and 3.

  7. Find a and b

    {a,b}={2,3}\{a,b\} = \{2,3\}

    So one log is 2 and the other is 3, in some order.

  8. Take the first case

    a=2, b=3:  log⁡2x=2,  log⁡2y=3a=2,\,b=3:\; \log_2 x = 2,\; \log_2 y = 3

    This assigns 2 to x and 3 to y.

  9. Convert to x and y

    x=22=4,y=23=8x = 2^2 = 4,\quad y = 2^3 = 8

    Convert each log back to a value.

  10. Take the second case

    a=3, b=2:  x=23=8,  y=22=4a=3,\,b=2:\; x = 2^3 = 8,\; y = 2^2 = 4

    Swapping the roles gives the second solution pair.

  11. Check the first pair

    log⁡24+log⁡28=2+3=5;(2)(3)=6  ✓\log_2 4+\log_2 8=2+3=5;\quad (2)(3)=6\;\checkmark

    For (4, 8): the logs are 2 and 3, which sum to 5 and multiply to 6, matching both equations.

  12. Check the second pair

    log⁡28+log⁡24=3+2=5;(3)(2)=6  ✓\log_2 8+\log_2 4=3+2=5;\quad (3)(2)=6\;\checkmark

    For (8, 4): the logs are 3 and 2, which again sum to 5 and multiply to 6.

  13. Explain why both are valid

    the equations are symmetric in x,y\text{the equations are symmetric in } x,y

    Swapping x and y leaves both equations unchanged, so if one pair works then the swapped pair works too.

  14. Note the key idea

    sum and product→quadratic in t\text{sum and product} \to \text{quadratic in } t

    Knowing the sum and product of two numbers let us build a quadratic whose roots are those numbers, a neat link back to quadratics.

  15. State the final answer

    (x,y)=(4,8)  or  (8,4)(x,y) = (4,8) \;\text{or}\; (8,4)

    Both ordered pairs are solutions of the simultaneous equations.

Answer
(x,y)=(4,8) or (8,4)(x,y) = (4,8) \text{ or } (8,4)
Question 2
8 markschallenging
Solve log⁡5(6⋅5x−5)=2x\log_5(6\cdot 5^{x} - 5) = 2x.
Show worked solution

Worked solution

  1. Convert the log to index form

    6⋅5x−5=5 2x6\cdot 5^x - 5 = 5^{\,2x}

    log_5(something) = 2x means that something equals 5 to the power 2x.

  2. Rewrite the right side as a square

    52x=(5x)25^{2x} = (5^x)^2

    By index laws, 5 to the power 2x is 5^x squared.

  3. Rewrite the equation

    6⋅5x−5=(5x)26\cdot 5^x - 5 = (5^x)^2

    Now every term is written using 5^x.

  4. Substitute a letter

    Let y=5x  (y>0)\text{Let } y = 5^x \;(y>0)

    This gives an ordinary quadratic.

  5. Rewrite in terms of y

    6y−5=y26y - 5 = y^2

    Replace 5^x with y.

  6. Form a quadratic

    y2−6y+5=0y^2 - 6y + 5 = 0

    Move all terms to one side.

  7. Factorise

    (y−1)(y−5)=0(y-1)(y-5) = 0

    Two numbers multiply to 5 and add to -6: they are -1 and -5.

  8. Solve for y

    y=1ory=5y = 1 \quad\text{or}\quad y = 5

    Both are positive; check the log's inside stays positive for both.

  9. Back-substitute the first

    5x=1=50  ⇒  x=05^x = 1 = 5^0 \;\Rightarrow\; x = 0

    For x = 0: 6(1) - 5 = 1 > 0, so the log is valid.

  10. Back-substitute the second

    5x=5=51  ⇒  x=15^x = 5 = 5^1 \;\Rightarrow\; x = 1

    For x = 1: 6(5) - 5 = 25 > 0, so the log is valid.

  11. Check x = 0

    log⁡5(6⋅50−5)=log⁡51=0=2(0)  ✓\log_5(6\cdot 5^0-5)=\log_5 1=0=2(0)\;\checkmark

    With x = 0: 6 - 5 = 1, and log_5 1 = 0, which matches 2x = 0.

  12. Check x = 1

    log⁡5(6⋅5−5)=log⁡525=2=2(1)  ✓\log_5(6\cdot 5-5)=\log_5 25=2=2(1)\;\checkmark

    With x = 1: 30 - 5 = 25, and log_5 25 = 2, which matches 2x = 2.

  13. Reflect on the method

    index form then a quadratic in 5x\text{index form then a quadratic in } 5^x

    Turning the log equation into index form exposed a hidden quadratic, which was the key idea here.

  14. Confirm both roots were kept

    y=1, 5 are both positivey=1,\,5 \text{ are both positive}

    Both y-values were positive, so neither solution had to be rejected.

  15. State the solutions

    x=0orx=1x = 0 \quad\text{or}\quad x = 1

    Both satisfy the original equation.

Answer
x=0 or x=1x = 0 \text{ or } x = 1
Question 3
8 markschallenging
Solve log⁡2(9−2x)=3−x\log_2(9 - 2^{x}) = 3 - x.
Show worked solution

Worked solution

  1. Convert the log to index form

    9−2x=2 3−x9 - 2^x = 2^{\,3-x}

    log_2(something) = 3 - x means that something equals 2 to the power (3 - x).

  2. Split the right-hand exponent

    23−x=232x=82x2^{3-x} = \dfrac{2^3}{2^x} = \dfrac{8}{2^x}

    Using index laws, 2^{3-x} is 8 divided by 2^x.

  3. Rewrite the equation

    9−2x=82x9 - 2^x = \dfrac{8}{2^x}

    Now both sides involve 2^x.

  4. Substitute a letter

    Let y=2x  (y>0)\text{Let } y = 2^x \;(y>0)

    This turns the equation into a rational equation in y.

  5. Rewrite in terms of y

    9−y=8y9 - y = \dfrac{8}{y}

    Replace 2^x with y throughout.

  6. Multiply through by y

    9y−y2=89y - y^2 = 8

    Clear the fraction by multiplying every term by y.

  7. Form a quadratic

    y2−9y+8=0y^2 - 9y + 8 = 0

    Rearrange into standard quadratic form.

  8. Factorise

    (y−1)(y−8)=0(y-1)(y-8) = 0

    Two numbers multiply to 8 and add to -9: they are -1 and -8.

  9. Solve for y

    y=1ory=8y = 1 \quad\text{or}\quad y = 8

    Both are positive, and both keep 9 - 2^x positive, so both are valid.

  10. Back-substitute the first

    2x=1=20  ⇒  x=02^x = 1 = 2^0 \;\Rightarrow\; x = 0

    1 is 2 to the power 0.

  11. Back-substitute the second

    2x=8=23  ⇒  x=32^x = 8 = 2^3 \;\Rightarrow\; x = 3

    8 is 2 cubed.

  12. Check x = 0

    log⁡2(9−20)=log⁡28=3=3−0  ✓\log_2(9-2^0)=\log_2 8=3=3-0\;\checkmark

    With x = 0: 9 - 1 = 8, and log_2 8 = 3, which equals 3 - x = 3.

  13. Check x = 3

    log⁡2(9−23)=log⁡21=0=3−3  ✓\log_2(9-2^3)=\log_2 1=0=3-3\;\checkmark

    With x = 3: 9 - 8 = 1, and log_2 1 = 0, which equals 3 - x = 0.

  14. Confirm the log stayed positive

    9−2x>0 for both roots9-2^x>0 \text{ for both roots}

    At x = 0 and x = 3 the inside 9 - 2^x is 8 and 1, both positive, so the logs are valid.

  15. State the solutions

    x=0orx=3x = 0 \quad\text{or}\quad x = 3

    Both satisfy the original equation.

Answer
x=0 or x=3x = 0 \text{ or } x = 3
Question 4
7 markschallenging
Given that y=log⁡2xy = \log_2 x, express log⁡2(8x4)−log⁡2x\log_2(8x^4) - \log_2 x in terms of yy, simplifying fully.
Show worked solution

Worked solution

  1. Split the first log with the product law

    log⁡2(8x4)=log⁡28+log⁡2(x4)\log_2(8x^4) = \log_2 8 + \log_2(x^4)

    A product inside a log becomes a sum of logs.

  2. Evaluate log_2 8

    log⁡28=3\log_2 8 = 3

    8 is 2 cubed, so its log base 2 is 3.

  3. Apply the power law to log_2(x^4)

    log⁡2(x4)=4log⁡2x=4y\log_2(x^4) = 4\log_2 x = 4y

    Bring the power 4 down; since y = log_2 x, this is 4y.

  4. Rewrite the whole expression

    (3+4y)−log⁡2x\big(3 + 4y\big) - \log_2 x

    Put the pieces of the first log together, then subtract the second log.

  5. Replace the last term

    =3+4y−y= 3 + 4y - y

    log_2 x is exactly y, so we subtract y.

  6. Simplify

    =3+3y= 3 + 3y

    4y minus y is 3y, giving the final simplified form.

  7. Recall the product law

    log⁡a(mn)=log⁡am+log⁡an\log_a(mn)=\log_a m+\log_a n

    This is why log_2(8x^4) split into log_2 8 plus log_2(x^4).

  8. Recall the power law

    log⁡a(mn)=nlog⁡am\log_a(m^n)=n\log_a m

    This is why the power 4 on x was allowed to come down to the front.

  9. Restate what y stands for

    y=log⁡2xy = \log_2 x

    Every log_2 x in the working becomes y, since that is how y was defined in the question.

  10. Rewrite log_2 8 as a number

    log⁡28=3\log_2 8 = 3

    8 is 2 cubed, so its log base 2 is exactly 3, a constant.

  11. Collect the pieces

    3+4y−y3 + 4y - y

    Bring together the constant 3, the 4y from the first log, and the -y from subtracting the second log.

  12. Combine the y terms

    4y−y=3y4y - y = 3y

    Subtracting y from 4y leaves 3y.

  13. Write the final simplified form

    3+3y3 + 3y

    So the whole expression equals 3 + 3y.

  14. Test with a value of x

    x=4⇒y=log⁡24=2x=4 \Rightarrow y=\log_2 4=2

    Pick x = 4 to test the answer; then y = 2.

  15. Verify the test

    log⁡2(8×44)−log⁡24=11−2=9=3+3(2)  ✓\log_2(8\times 4^4)-\log_2 4 = 11 - 2 = 9 = 3+3(2)\;\checkmark

    Directly, log_2(8 x 256) = log_2 2048 = 11, minus 2 gives 9; and 3 + 3(2) = 9, so the formula works.

Answer
3+3y3 + 3y
Question 5
8 markschallenging
Solve 3x+32−x=103^{x} + 3^{2-x} = 10.
Show worked solution

Worked solution

  1. Rewrite the second term

    32−x=323x=93x3^{2-x} = \dfrac{3^2}{3^x} = \dfrac{9}{3^x}

    Using index laws, 3^{2-x} is 9 divided by 3^x.

  2. Rewrite the equation

    3x+93x=103^x + \dfrac{9}{3^x} = 10

    Now everything is in terms of 3^x.

  3. Substitute a letter

    Let y=3x  (y>0)\text{Let } y = 3^x \;(y>0)

    This gives a neater equation; y must be positive.

  4. Rewrite in terms of y

    y+9y=10y + \dfrac{9}{y} = 10

    Replace 3^x with y throughout.

  5. Multiply through by y

    y2+9=10yy^2 + 9 = 10y

    Clear the fraction by multiplying by y.

  6. Form a quadratic

    y2−10y+9=0y^2 - 10y + 9 = 0

    Bring all terms to one side.

  7. Factorise

    (y−1)(y−9)=0(y-1)(y-9) = 0

    Two numbers multiply to 9 and add to -10: they are -1 and -9.

  8. Solve for y

    y=1ory=9y = 1 \quad\text{or}\quad y = 9

    Both are positive and valid.

  9. Back-substitute the first

    3x=1=30  ⇒  x=03^x = 1 = 3^0 \;\Rightarrow\; x = 0

    1 is 3 to the power 0.

  10. Back-substitute the second

    3x=9=32  ⇒  x=23^x = 9 = 3^2 \;\Rightarrow\; x = 2

    9 is 3 squared.

  11. Check x = 0

    30+32=1+9=10  ✓3^{0} + 3^{2} = 1 + 9 = 10\;\checkmark

    With x = 0: 3^x=1 and 3^{2-x}=3^2=9, which sum to 10.

  12. Check x = 2

    32+30=9+1=10  ✓3^{2} + 3^{0} = 9 + 1 = 10\;\checkmark

    With x = 2: 3^x=9 and 3^{2-x}=3^0=1, which sum to 10.

  13. Note the symmetry

    x↔2−x swaps the termsx \leftrightarrow 2-x \text{ swaps the terms}

    Replacing x with 2 - x swaps the two terms, so the roots come as a matching pair about x = 1.

  14. Confirm both roots were kept

    y=1, 9 are both positivey=1,\,9 \text{ are both positive}

    Both y-values were positive, so neither solution had to be rejected.

  15. State the solutions

    x=0orx=2x = 0 \quad\text{or}\quad x = 2

    Both satisfy the original equation.

Answer
x=0 or x=2x = 0 \text{ or } x = 2

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