Exponential growth and decay are modelled by P = Aeᵏᵗ. A is the starting value, when t = 0. If k > 0 the quantity grows; if k < 0 it decays.
To find k from data, substitute a known value and use ln. If P = 850eᵏᵗ and P = 1200 when t = 3, then 1200/850 = e³ᵏ, so k = ln(1200/850) ÷ 3 = 0.115. Equations like 3²ˣ + 5 × 3ˣ − 6 = 0 are quadratics in 3ˣ: (3ˣ + 6)(3ˣ − 1) = 0 gives 3ˣ = 1, so x = 0.
Worked example
Solve 5ˣ⁺² = 7ˣ to 3 s.f.
- Take logs: (x + 2) ln 5 = x ln 7.
- x(ln 7 − ln 5) = 2 ln 5, so x = 2 ln 5 ÷ ln 1.4 = 9.565…
Answer:x = 9.57
Key words
- exponential model
- an equation like P = Aeᵏᵗ
- growth constant (k)
- the rate in the exponent
- e
- about 2.718, the natural base
Practise this
New questions every time.
Common Core-referenced: HSF-LE.A.4