Exponential and logarithmic functions; solving equations

Maths: Analysis and Approaches · Functions. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

An exponential function y = a^x has the variable in the power. Its graph passes through (0, 1), is always above the x-axis and has the horizontal asymptote y = 0. It rises if a > 1 and falls if 0 < a < 1.

The logarithmic function y = log_a x is its inverse, so its graph is the reflection in y = x. It passes through (1, 0), exists only for x > 0 and has the vertical asymptote x = 0.

Growth and decay models have the form A = A₀e^kt. To find a time, isolate the exponential and take ln. Some equations have no algebraic method, so use a graph.

Rules to remember

y = a^x: passes through (0, 1); asymptote y = 0; domain all real x; range y > 0
y = log_a x: passes through (1, 0) and (a, 1); asymptote x = 0; domain x > 0; range all real y
Inverses: log_a(a^x) = x and a^{log_a x} = x; ln(e^x) = x and e^{ln x} = x
y = e^x + c has horizontal asymptote y = c
A = A₀e^kt: A₀ is the value at t = 0; k > 0 growth, k < 0 decay
Doubling time = (ln 2)/k for growth e^kt; half-life = (ln 2)/k for decay e^−kt
After solving a log equation, reject any answer that makes the inside of a log zero or negative

Common mistake

Keeping every root of a log equation. Substitute each answer back: the logarithm of zero or of a negative number does not exist, so those answers are rejected.

Worked example

The mass of a substance is A = 120e^−0.05t grams, with t in years. Find the time for the mass to fall to 30 g.

30 = 120e^−0.05t, so e^−0.05t = 30 ÷ 120 = 0.25.
Take ln: −0.05t = ln 0.25 = −ln 4.
t = (ln 4) ÷ 0.05 = 20 ln 4 = 20 × 1.3863.

Answer: t = 20 ln 4 ≈ 27.7 years

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