Example 1: An exponential equation that is a quadratic in eˣ
Solve e²ˣ − 3eˣ + 2 = 0.
- Let
u = eˣ:u² − 3u + 2 = 0 (u − 1)(u − 2) = 0eˣ = 1oreˣ = 2x = 0orx = ln 2
Answer: x = 0 or x = ln 2
e ≈ 2.71828 is the limit of (1 + 1/n)ⁿ as n tends to infinity. The natural logarithm ln x is the logarithm to base e, so ln eˣ = x and eln x = x. e matters in M2 because eˣ is its own derivative and ln x has derivative 1/x.
e = limn→∞ (1 + 1/n)ⁿeˣ = 1 + x + x²/2! + x³/3! + …ln eˣ = x, eln x = x (x > 0)aˣ = ex ln aloga x = ln x / ln aln(xy) = ln x + ln y, ln(x/y) = ln x − ln y, ln xⁿ = n ln xSolve e²ˣ − 3eˣ + 2 = 0.
u = eˣ: u² − 3u + 2 = 0(u − 1)(u − 2) = 0eˣ = 1 or eˣ = 2x = 0 or x = ln 2Answer: x = 0 or x = ln 2
Find dy/dx if y = log₂ x.
y = ln x / ln 21/ln 2 is a constant, so dy/dx = (1/ln 2) · (1/x)Answer: dy/dx = 1 / (x ln 2)
eˣ = −3, reject it: eˣ is always positive. Likewise, check that every solution keeps the argument of each ln positive.ln(x + 2) cannot be split.How to study: Click any question below to instantly load it on the coordinate grid and check your steps.