Logarithmic and exponential functions

Maths Pure 3 · Logarithmic and exponential 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

A logarithm is a power. The statement log_a x = y means exactly the same as a^y = x. The natural logarithm ln x uses the base e ≈ 2.718, so ln x and e^x undo each other.

The graph of y = e^x passes through (0, 1), is always above the x-axis and rises more and more steeply. The graph of y = ln x is its reflection in the line y = x: it passes through (1, 0) and exists only for x > 0.

To solve an equation with the unknown in a power, take logarithms of both sides. Logarithms also turn the curves y = kxⁿ and y = k(a^x) into straight lines.

Rules to remember

log_a x = y means a^y = x; so log_a 1 = 0 and log_a a = 1
log x + log y = log(xy); log x − log y = log(x/y); log(xⁿ) = n log x
ln(e^x) = x and e^{ln x} = x (for x > 0)
a^x = b gives x = ln b ÷ ln a
In an inequality, dividing by ln a reverses the sign when 0 < a < 1, because ln a is then negative
y = kxⁿ gives ln y = n ln x + ln k: plot ln y against ln x; gradient = n, intercept = ln k
y = k(a^x) gives ln y = (ln a)x + ln k: plot ln y against x; gradient = ln a, intercept = ln k

Common mistake

Students write ln(x + y) as ln x + ln y. The laws only work for products, quotients and powers: ln x + ln y = ln(xy). Also remember that the intercept of the straight line is ln k, not k, so k = e^intercept.

Worked example

The variables x and y satisfy y = kxⁿ. The graph of ln y against ln x is a straight line through (2, 5.0) and (6, 7.0). Find n and k.

ln y = n ln x + ln k, so the gradient is n: n = (7.0 − 5.0) ÷ (6 − 2) = 0.5.
Intercept: ln k = 5.0 − 0.5 × 2 = 4.0.
k = e^4.0 = 54.6 (3 significant figures).

Answer: n = 0.5 and k = 54.6

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