Questions on logarithms and their laws

Question 1

Write down the values of log₂ 32, log₅ 125 and log₁₀ 0.01.

Accepted Answer

A logarithm is an exponent. Since 2⁵ = 32 the first is 5, since 5³ = 125 the second is 3, and since 10⁻² = 0.01 the third is −2. B1 B1 B1, one for each value. Logs of numbers between 0 and 1 are negative rather than undefined.

Question 2

Write log 8 + log 5 − log 10 as a single logarithm.

Accepted Answer

Combine left to right. log 8 + log 5 = log 40, then log 40 − log 10 = log (40/10) = log 4. M1 for one correct use of a log law, A1 for log 4. Writing log 8 + log 5 as log 13 is wrong.

Question 3

Write the statement 6³ = 216 as a logarithmic statement, write log₄ 64 = 3 as an exponential statement, and write e^x = 20 as a logarithmic statement using ln.

Accepted Answer

6³ = 216 becomes log₆ 216 = 3: the base of the power stays the base of the logarithm, and the logarithm names the exponent. log₄ 64 = 3 becomes 4³ = 64, the same swap in the other direction. e^x = 20 becomes x = ln 20, since ln is the logarithm to base e. B1 B1 B1, one for each conversion.

Question 4

Explain why the definition of log_a x requires a > 0 and a ≠ 1, and why x must be positive.

Accepted Answer

aⁿ is only guaranteed to be a single, unambiguous real number for every real n when a > 0, and a ≠ 1 is needed because 1ⁿ = 1 for every n, so log₁ x could not name an exponent for any x other than 1, and no single one for x = 1 itself. B1 for both base conditions. Since a > 0, aⁿ is positive for every real n, so no power of a ever produces zero or a negative number: x must be positive for log_a x to name an exponent at all. B1 for the positivity of x.

Question 5

Write ½ log 49 + log 2 as a single logarithm.

Accepted Answer

The multiple moves inside first, so ½ log 49 = log 49^1/2 = log 7. Then log 7 + log 2 = log 14. M1 for taking the multiple inside, A1 for log 7, A1 for log 14. Doing the addition before the halving gives ½ log 98, which is a different number.

Question 6

Solve e^2x−1 = 12, giving your answer to 4 significant figures.

Accepted Answer

Take ln of both sides to get 2x − 1 = ln 12, so x = (ln 12 + 1)/2 = 1.742 to 4 significant figures. M1 for taking logs, A1 for 2x − 1 = ln 12, A1 for 1.742. Keep ln 12 exact until the last line; rounding it to 2.48 first can shift the fourth figure.

Question 7

Solve ln (2x + 5) = 3, giving your answer to 4 significant figures.

Accepted Answer

Apply e to both sides, giving 2x + 5 = e³, so x = (e³ − 5)/2 = 7.543 to 4 significant figures. M1 for applying e to both sides, A1 for 2x + 5 = e³, A1 for 7.543. Note that e³ undoes the whole ln bracket at once; “e³ = 2x, then + 5” is the misstep.

Question 8

Solve 5^2x+1 = 40, giving your answer to 4 significant figures.

Accepted Answer

The bases refuse to match, so take logs of both sides and bring the power down: (2x + 1) log 5 = log 40. Then log 40/log 5 = 2.292, so 2x + 1 = 2.292 and x = 0.6460 to 4 significant figures. M1 for taking logs, A1 for (2x + 1) log 5 = log 40, M1 for rearranging, A1 for 0.6460. Any base of logarithm serves, because it cancels in the ratio. What does not work is log 40/log 5 = log 8; a quotient of logs is not the log of a quotient.

Question 9

Expand log (8x³/y) as a sum and difference of simpler logarithms, writing your answer using log 2, log x and log y, given x, y > 0.

Accepted Answer

First split the quotient: log (8x³/y) = log (8x³) − log y. M1 for the quotient law. Then split the product and bring down the power: log (8x³) = log 8 + log x³ = 3 log 2 + 3 log x, since 8 = 2³. M1 for the product and power laws together. So the expansion is 3 log 2 + 3 log x − log y. A1 for the complete expression.

Question 10

Solve log₃(2x − 1) = 4.

Accepted Answer

Convert to the exponential statement 2x − 1 = 3⁴ = 81. M1 for the conversion. Then 2x = 82, so x = 41. A1 for 2x = 82, A1 for x = 41.

Question 11

Solve log x + log (x − 3) = 1.

Accepted Answer

Combine using the first law: log [x(x − 3)] = 1, so x(x − 3) = 10¹ = 10. M1 for combining the logs, A1 for x² − 3x − 10 = 0. Factorising, (x − 5)(x + 2) = 0, so x = 5 or x = −2. M1 for solving the quadratic. Both are roots of the quadratic, but log (x − 3) needs x > 3, and x = −2 fails that as well as making log x undefined, so it is rejected: x = 5. A1 for rejecting x = −2 and giving x = 5.

Question 12

Solve 2^2x − 5 × 2^x + 4 = 0.

Accepted Answer

Let u = 2^x; since 2^2x = u², the equation is u² − 5u + 4 = 0, which factorises as (u − 1)(u − 4) = 0. So 2^x = 1, giving x = 0, or 2^x = 4, giving x = 2. M1 for the substitution, A1 for the quadratic in u, M1 for solving it, A1 A1 for the two values of x.

Question 13

Solve (log x)² − log (x³) − 4 = 0.

Accepted Answer

Let L = log x, so log (x³) = 3L and the equation becomes L² − 3L − 4 = 0. M1 for the substitution, A1 for the quadratic in L. This factorises as (L − 4)(L + 1) = 0, so L = 4 or L = −1. M1 for solving it. L = 4 gives log x = 4, so x = 10⁴ = 10000; L = −1 gives log x = −1, so x = 10⁻¹ = 0.1. A1 A1 for the two values of x. Both are positive, so both are valid.

Question 14

Solve the simultaneous equations log x + log y = log 12 and 2 log x − log y = log 18, for x and y, both positive.

Accepted Answer

Combine each side with the log laws: the first equation is log (xy) = log 12, so xy = 12, and the second is log (x²/y) = log 18, so x²/y = 18. M1 for combining each equation. Adding the two original equations directly avoids handling x and y separately: (log x + log y) + (2 log x − log y) = 3 log x = log 12 + log 18 = log 216. M1 for the addition. Since 216 = 6³ exactly, 3 log x = 3 log 6, so x = 6. M1 for taking the cube root via the exponent, A1 for x = 6. Substituting into xy = 12 gives y = 2. A1 for y = 2. A quadratic in x reached by substitution would work as well, but combining the two given equations directly by addition removes y in one line.

Question 15

The pH of a solution is defined by pH = −log₁₀ [H⁺], where [H⁺] is the hydrogen-ion concentration in mol/L. Solution A has pH 3 and solution B has pH 5. Find [H⁺] for each solution as a power of 10, find the ratio [H⁺]_A : [H⁺]_B, and explain why solution A is 100 times more acidic than solution B rather than only 2 units more acidic.

Accepted Answer

Converting the definition to exponential form, [H⁺] = 10^−pH. So solution A has [H⁺]_A = 10⁻³ mol/L and solution B has [H⁺]_B = 10⁻⁵ mol/L. B1 B1 for the two concentrations. The ratio is [H⁺]_A / [H⁺]_B = 10⁻³/10⁻⁵ = 10² = 100. M1 for forming the ratio, A1 for 100. The pH is a logarithm of the concentration, and a fixed difference in a logarithm corresponds to a fixed ratio, not a fixed difference, in the quantity itself: subtracting 2 from the exponent −3 to reach −5 divides the concentration by 10², which is why 2 pH units apart means 100 times the concentration, not twice it. B1 for the explanation in terms of the exponent.