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 and evaluate.

• 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.