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Logarithms and their laws questions

A logarithm answers one question. What power produced this number? That single idea undoes every exponential, drags unknowns down out of exponents where algebra cannot otherwise reach them, and obeys three laws that turn multiplication into addition.

7 original questions · 23 marks · the logarithms and their laws notes · Exponentials and logarithms

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

3 marks · write · warm-up
Write down the values of log₂ 32, log₅ 125 and log₁₀ 0.01.
Worked 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. That last one trips people. Logs of numbers below 1 are negative, not impossible, and writing “undefined” loses the mark outright.
2 marks · write · warm-up
Write log 8 + log 5 − log 10 as a single logarithm.
Worked 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. Addition multiplies and subtraction divides, which is the index laws wearing log clothing. Writing log 8 + log 5 as log 13 is the error the first law exists to kill.
3 marks · write
Write ½ log 49 + log 2 as a single logarithm.
Worked 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. The third law handles fractional multipliers exactly as it handles whole ones, and a half becomes a square root. Doing the addition before the halving gives ½ log 98, which is a different number.
3 marks
Solve e^2x−1 = 12, giving your answer to 4 significant figures.
Worked 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. The log peels the exponential off, and everything after that is linear algebra. Keep ln 12 exact until the last line; rounding it to 2.48 first can shift the fourth figure.
3 marks
Solve ln (2x + 5) = 3, giving your answer to 4 significant figures.
Worked 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. Here the exponential has to be applied rather than peeled off, and deciding which way round is the real skill. Note that e³ undoes the whole ln bracket at once; “e³ = 2x, then + 5” is the misstep.
4 marks
Solve 5^2x+1 = 40, giving your answer to 4 significant figures.
Worked 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.
5 marks · stretch
Solve 2^2x − 5 × 2^x + 4 = 0.
Worked 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. A quadratic wearing an exponential disguise: the substitution unlocks it, and both roots here convert back cleanly.

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