Mathematics: TestAS practice testlets

A quantity grows exponentially if it is multiplied by the same factor in equal time intervals. A bacterial culture in which every cell divides every 20 minutes doubles its size every 20 minutes. Starting from 500 bacteria, it contains 1,000 after 20 minutes, 2,000 after 40 minutes and 4,000 after one hour. The function describing this growth has the variable in the exponent.

N(t) = N₀·2^(t ÷ T_d) N(t) = N₀·aᵗ solving aᵗ = b: t = log(b) ÷ log(a) 2¹⁰ = 1,024

Exponential growth is at first slow and then extremely fast. Each doubling adds as much as all previous growth together. This is why the early phase of an epidemic or of a chain reaction is easily underestimated. In reality, exponential growth never continues indefinitely: bacteria run out of nutrients and space, and the growth slows down, often following a logistic curve that levels off at a maximum. The same model with a limit describes the spread of new technologies and the growth of animal populations in a habitat of limited size.

Exponential decay occurs when a quantity is multiplied by a factor smaller than one in equal time intervals. If the body eliminates 20 percent of a drug every hour, 80 percent remains after each hour, and the amount after t hours is the initial amount times 0.8ᵗ. After three hours, 0.8 × 0.8 × 0.8 = 0.512, or 51.2 percent, remains. The time after which half remains is the half-life.

To find the time at which an exponential function reaches a given value, the equation must be solved for the exponent, which requires logarithms. The logarithm answers the question: to which power must the base be raised to give a certain number? Because 2¹⁰ = 1,024, the logarithm to base 2 of about 1,000 is about 10: a quantity that doubles must double about ten times to increase a thousandfold. On a logarithmic scale, exponential growth appears as a straight line, which makes it easy to recognise in measured data. Logarithmic scales are therefore used for the number of cases in epidemics, for earthquake magnitudes and for the loudness of sounds in decibels. A calculator or the table of powers of two is enough to estimate most times of growth and decay without a formal logarithm. The questions refer to the bacterial culture and the drug described above.

Some formulas contain the unknown in an exponent or in a reciprocal, and rearranging them requires special steps. The compound interest formula K = K₀·(1 + p)ⁿ gives the final capital K after n years at an interest rate p, written as a decimal. Three different questions can be asked of it, each requiring a different rearrangement.

To find the final capital, the formula is used directly. To find the interest rate that turns K₀ into K in n years, one first divides by K₀, giving (1 + p)ⁿ = K ÷ K₀, then takes the n-th root, giving 1 + p = (K ÷ K₀)^(1/n), and finally subtracts 1. To find the number of years, the unknown stands in the exponent, and logarithms are needed: taking the logarithm of both sides of (1 + p)ⁿ = K ÷ K₀ gives n·log(1 + p) = log(K ÷ K₀), so n = log(K ÷ K₀) ÷ log(1 + p).

K = K₀·(1 + p)ⁿ p = (K ÷ K₀)^(1/n) − 1 n = log(K ÷ K₀) ÷ log(1 + p) 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ ⇒ R = R₁·R₂ ÷ (R₁ + R₂)

Reciprocal formulas occur for resistors in parallel, lenses and many rates. For two resistors in parallel, 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂. Adding the fractions on the right gives (R₁ + R₂) ÷ (R₁·R₂), and taking the reciprocal of both sides gives R = R₁·R₂ ÷ (R₁ + R₂). To find the second resistor needed for a desired total, one isolates its reciprocal first: 1 ÷ R₂ = 1 ÷ R − 1 ÷ R₁ = (R₁ − R) ÷ (R·R₁), so R₂ = R·R₁ ÷ (R₁ − R). A common error is to take the reciprocal of each term separately, which is not allowed: the reciprocal of a sum is not the sum of the reciprocals.

As always, the result should be checked by inserting it into the original formula and by asking whether it is plausible. The total resistance of two resistors in parallel, for example, must always be smaller than each of the individual resistors, because the current has two paths instead of one.