Exponentials & Logarithms

Solve the equation log₂ x = 5 for x.

• A32✓
• B16
• C10
• D25

Answer: By definition log₂ x = 5 means x = 2⁵ = 32.

Write log₁₀ 8 + log₁₀ 5 as a single logarithm.

• Alog₁₀ 40✓
• Blog₁₀ 1.6
• Clog₁₀ 13
• Dlog₁₀ 3

Answer: By the addition law of logs, log 8 + log 5 = log(8 × 5) = log₁₀ 40.

Solve 3^x = 81 for x.

• A9
• B3
• C4✓
• D27

Answer: Since 81 = 3⁴, the equation 3^x = 3⁴ gives x = 4.

Given ln x = 0, find the value of x.

• A0
• Be
• C1✓
• D−1

Answer: ln x = 0 means x = e⁰ = 1.

Write 2 log x − log 3 as a single logarithm (same base, x > 0).

• Alog(2x/3)
• Blog(x² − 3)
• Clog(x²/3)✓
• Dlog(2x − 3)

Answer: By the power law 2 log x = log x², and by the subtraction law log x² − log 3 = log(x²/3). The distractor log(2x/3) wrongly treats the coefficient 2 as a multiplier.

Solve 5ˣ = 20, giving your answer to 3 significant figures.

• A4.00
• B1.30
• C0.537
• D1.86✓

Answer: Taking logs, x = log 20 / log 5 = 1.301/0.699 = 1.86 (3 s.f.). The distractor 4.00 comes from dividing 20/5; 1.30 is log₁₀ 20 alone.

Solve e^(2x) = 7, giving your answer as an exact value.

• Aln 7
• B2 ln 7
• C(ln 7)/2✓
• Dln 3.5

Answer: Taking natural logs of both sides: 2x = ln 7, so x = (ln 7)/2. The distractor ln 7 forgets to divide by 2; ln 3.5 comes from wrongly computing ln(7/2).

A radioactive sample has mass m = 50e^(−0.02t) grams, where t is the time in years. Find the mass after 100 years, to 3 significant figures.

• A6.77 g✓
• B0.135 g
• C369 g
• D7.00 g

Answer: At t = 100, m = 50e^(−0.02 × 100) = 50e^(−2) = 50 × 0.1353 = 6.77 g. The distractor 0.135 g omits the factor 50; 369 g comes from a sign error giving 50e^(+2).

A population is modelled by P = 2000e^(0.05t), where t is the time in years. Find the value of t when P = 5000, to 3 significant figures.

• A50.0
• B18.3✓
• C7.96
• D25.0

Answer: 5000 = 2000e^(0.05t) gives e^(0.05t) = 2.5, so 0.05t = ln 2.5 = 0.9163 and t = 18.3 years. The distractor 7.96 uses log₁₀ instead of ln; 50.0 comes from 2.5/0.05.

A relationship of the form y = ax^n is investigated by plotting log₁₀ y against log₁₀ x. The result is a straight line with gradient 1.5 and vertical-axis intercept 0.6. Find the values of n and a.

• An = 0.6, a = 31.6
• Bn = 1.5, a = 0.6
• Cn = 3.98, a = 1.5
• Dn = 1.5, a = 3.98✓

Answer: Taking logs of y = ax^n gives log y = n log x + log a, a straight line in log x. So the gradient is n = 1.5 and the intercept is log a = 0.6, giving a = 10^0.6 = 3.98. The distractor n = 1.5, a = 0.6 forgets to undo the log on the intercept.

Solve 2^x = 32.

• Ax = 5✓
• Bx = 16
• Cx = 1/5
• Dx = 32

Answer: Ask what power of 2 gives 32: 2⁵ = 32, so x = 5. Taking logs of both sides gives x = log2(32) = 5, which is the method that keeps working when the answer is not a whole number.

Solve log₂(x) + log₂(3) = 5.

• Ax = 32/3✓
• Bx = 96
• Cx = 29/3
• Dx = 32

Answer: Combine the logs first: log₂(x) + log₂(3) = log₂(3x). So log₂(3x) = 5 gives 3x = 2⁵ = 32 and x = 32/3. Multiplying by 3 instead of dividing gives 96, and subtracting 3 from 32 treats the sum of two logs as a sum of two numbers.

Solve 5^x = 125.

• Ax = 125
• Bx = 25
• Cx = 1/3
• Dx = 3✓

Answer: Ask what power of 5 gives 125: 5³ = 125, so x = 3. Taking logs of both sides gives x = log5(125) = 3, which is the method that keeps working when the answer is not a whole number.

A population is modelled by P = 500e^(t/10), where t is the time in years. Find P when t = 20, to 3 s.f.

• A502
• B553
• C1000
• D3690✓

Answer: Substitute t = 20: P = 500e^(20/10) = 500e² = 3694.5..., which is 3690 to 3 s.f. The exponent must be evaluated before the exponential; multiplying 500 by the rate is a linear-model error, not an exponential one.

A population is modelled by P = 200e^(t/5), where t is the time in years. Find P when t = 15, to 3 s.f.

• A4020✓
• B244
• C600
• D203

Answer: Substitute t = 15: P = 200e^(15/5) = 200e³ = 4017.1..., which is 4020 to 3 s.f. The exponent must be evaluated before the exponential; multiplying 200 by the rate is a linear-model error, not an exponential one.

Evaluate log_2(8) + log_2(2).

• A3
• B4✓
• C10
• D5

Answer: log_2(8) = 3 and log_2(2) = 1, so the sum is 4. Equivalently, adding logs multiplies the arguments: log_2(16) = 4.

Evaluate log_3(27) + log_3(3).

• A3
• B4✓
• C30
• D5

Answer: log_3(27) = 3 and log_3(3) = 1, so the sum is 4. Equivalently, adding logs multiplies the arguments: log_3(81) = 4.

Write 2log(x) − log(y) as a single logarithm.

• Alog(x²/y)✓
• Blog(2x/y)
• Clog(x²y)
• Dlog(2x − y)

Answer: The coefficient becomes a power: 2log(x) = log(x²). Subtracting logs divides the arguments, so log(x²) − log(y) = log(x²/y). Moving a coefficient in front of a log to multiply the argument is the most common log-law slip.

What is log₁₀(1)?

• A1, because any logarithm of its own base returns the value one
• B0✓
• C10
• DUndefined, because you cannot take the logarithm of one

Answer: Ask what power of 10 gives 1. Since 10⁰ = 1, the answer is 0. This holds in any base — log_b(1) = 0 always, because anything to the power zero is one.

Why is log(−5) undefined for real numbers?

• ABecause the calculator cannot display the answer to sufficient accuracy
• BBecause logarithms are only defined for whole numbers greater than one
• CNo real power of a positive base gives a negative result✓
• DBecause negative numbers have two different logarithms rather than one

Answer: Raising a positive base to any real power always gives a positive result, so nothing can produce −5. That is also why the graph of log x lies entirely to the right of the y-axis and has the y-axis as an asymptote.

The number of bacteria in a culture is modelled by N = 400e^(kt), where t is in hours. After 3 hours there are 1000 bacteria. Find k, to 3 significant figures.

• Ak = 0.135
• Bk = 0.833
• Ck = 2.75
• Dk = 0.305✓

Answer: 1000 = 400e^(3k), so e^(3k) = 2.5 and 3k = ln 2.5, giving k = 0.916/3 = 0.305. Using ln(600/400) treats the growth as an increase of 600 compared with the start; k = 0.833 divides 2.5 by 3 without taking the logarithm.

After a dose, the concentration of a drug in the blood decays as 8e^(−0.15t) mg per litre, t hours after the dose. A patient takes one dose at t = 0 and an identical dose at t = 6; the two contributions add. How long after the first dose does the concentration first fall back below 3 mg per litre after the second dose, to 3 significant figures?

• A12.5 hours
• B8.81 hours
• C6.54 hours
• D14.8 hours✓

Answer: For t ≥ 6, C = 8e^(−0.15t) + 8e^(−0.15(t − 6)) = 8e^(−0.15t)(1 + e^(0.9)). Setting C = 3: e^(−0.15t) = 3/(8 × 3.4596), so t = ln(9.2256)/0.15 = 14.8 hours. The value 8.81 hours is the time after the second dose, not after the first; 12.5 hours ignores the drug left from the first dose.

A quantity P is modelled by P = Ae^(kt). A graph of ln P against t is a straight line passing through (0, 4.2) and (10, 5.7). Find A and k, to 3 significant figures.

• AA = 66.7, k = 0.150✓
• BA = 4.20, k = 0.150
• CA = 66.7, k = 1.16
• DA = 15 800, k = 0.150

Answer: Taking logs: ln P = ln A + kt, a line with gradient k and intercept ln A. Gradient = (5.7 − 4.2)/10 = 0.15, so k = 0.150; ln A = 4.2, so A = e^4.2 = 66.7. A = 4.20 reads the intercept as A itself, and 15 800 uses 10^4.2, which would be right only for log₁₀.