Exponentials and logarithms flashcards

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Exponential function
A function y = aˣ where the base a is a positive constant, a ≠ 1, and the variable x is the power.
Logarithm to base a of b
The power to which a must be raised to give b: log_a b = x means aˣ = b (a > 0, a ≠ 1, b > 0).
Asymptote of y = aˣ
The x-axis, y = 0: the curve gets ever closer to it but never meets it, because aˣ > 0 for all x.
log on a calculator (P2)
log with no base written means log₁₀. (The natural log ln, base e, is P3.)
Domain of log_a x
x > 0 only: there is no logarithm of zero or of a negative number.
Product law of logarithms
log_a(xy) = log_a x + log_a y (No — must be known)
Quotient law of logarithms
log_a(x/y) = log_a x − log_a y (No — must be known)
Power law of logarithms
log_a(xᵏ) = k log_a x (No — must be known)
Reciprocal law of logarithms
log_a(1/x) = −log_a x (No — must be known)
log_a a and log_a 1
log_a a = 1 (since a¹ = a); log_a 1 = 0 (since a⁰ = 1). (No — must be known)
Change of base formula
log_a x = log_b x ÷ log_b a (Yes — in the formulae booklet)
Solution of aˣ = b
x = log b ÷ log a (= log_a b), valid for b > 0. (No — derive it)
Key features of y = aˣ, a > 1
Increasing; passes through (0, 1) and (1, a); x-axis is an asymptote as x → −∞; always above the x-axis.
Key features of y = aˣ, 0 < a < 1
Decreasing; passes through (0, 1); x-axis is an asymptote as x → +∞; it is the reflection of y = (1/a)ˣ in the y-axis.
Graph of y = log_a x compared with y = aˣ (a > 1)
Reflection in y = x: passes through (1, 0) and (a, 1), defined only for x > 0, with the y-axis as an asymptote.
Effect of +k on y = aˣ + k
Translation by k upwards: the y-intercept becomes 1 + k and the asymptote becomes y = k.
Solving aˣ = bˣ⁺ᶜ graphically
The x-coordinate where the curves y = aˣ and y = bˣ⁺ᶜ cross.
Why the laws of logs hold
They are index laws read backwards: if log_a x = m and log_a y = n then xy = aᵐ⁺ⁿ, so log_a(xy) = m + n.
Spotting a quadratic in aˣ
Terms in a²ˣ (or 4ˣ, 9ˣ…) and aˣ together: let u = aˣ, so a²ˣ = u², and solve the quadratic in u.
Why u = aˣ must be positive
aˣ > 0 for all real x, so any root u ≤ 0 gives no solution and is rejected with that reason.
Method for a log equation
Same base → combine into one log (each side) → remove logs → solve → check every argument is positive.
log_a b in terms of log_b a
log_a b = 1 ÷ log_b a (change of base with base b).
log_(a²) x in terms of log_a x
½ log_a x, since log_a(a²) = 2.
Find log₄ 8 exactly.
4ʸ = 8 → 2²ʸ = 2³ → y = 3/2.
Simplify 2 log₃ 6 − log₃ 4.
log₃ 36 − log₃ 4 = log₃ 9 = 2.
Solve 5ˣ = 40 to 3 s.f.
x = log 40 ÷ log 5 = 2.29.
Solve 2ˣ⁺¹ = 3ˣ to 3 s.f.
(x + 1) log 2 = x log 3 → x = log 2 ÷ (log 3 − log 2) = 1.71.
Solve 4ˣ − 5 × 2ˣ − 6 = 0.
u = 2ˣ: (u − 6)(u + 1) = 0; u = −1 impossible; 2ˣ = 6 → x = 2.58 (3 s.f.).
Solve 2 log₃(x − 1) − log₃(x + 5) = 1, x > 1.
(x − 1)² = 3(x + 5) → x² − 5x − 14 = 0 → x = 7 (x = −2 rejected: log₃(x − 1) undefined).
Solve log₃ x = 4 logₓ 3.
L = log₃ x: L = 4/L → L² = 4 → L = ±2 → x = 9 or 1/9.
log(x + y) — can it be split?
No. Only log(xy) splits into a sum; log(x + y) has no law.
log 8 ÷ log 2 versus log(8 ÷ 2)
log 8 ÷ log 2 = log₂ 8 = 3; log(8 ÷ 2) = log 4. They are different.
Rejecting roots of a log equation
Reject a root only if it makes some log argument ≤ 0 in the original equation — a negative x itself can be fine.
Rounding in exponential equations
Keep the exact form (e.g. log 40 ÷ log 5) until the last line; round only the final answer, to 3 s.f. unless told otherwise.
Rewriting aˣ⁺ᵏ before substituting
aˣ⁺ᵏ = aᵏ × aˣ (e.g. 3ˣ⁺¹ = 3 × 3ˣ), not aˣ + k.

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