The logarithmic function y = logₐ x is the inverse of the exponential function y = aˣ.
The exponential function answers "what do you get by multiplying a x times?", while the logarithmic function answers "how many times must a be multiplied to get x?", so the two graphs face each other across the line y = x.
That is why a logarithmic graph always passes through the points (1, 0) and (a, 1), and drops without bound as x approaches 0, with the y-axis as its asymptote.
In the first picture, changing the base moves the point (a, 1) and changes how flat the curve lies; the second picture marks two pairs of points facing each other across y = x.
How the shape depends on the base
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The curve y = logₐ x for the chosen base (gold), the curves for bases 2, 3, 10 (faint), and the points (1, 0) and (a, 1)
2
💡 Every logarithmic graph passes through (1, 0)
①Since logₐ 1 = 0, the graph passes through (1, 0) whatever the base
②Since logₐ a = 1, it also passes through (a, 1); the bigger the base, the further right this point sits and the flatter the curve
③As x gets close to 0, y decreases without bound, so the y-axis is an asymptote
④The base in this picture is at least 1.5, so only increasing curves are drawn
Symmetry with the exponential function
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y = 2ˣ (red) and y = log₂ x (gold), with the points (0, 1) and (1, 0), and (1, 2) and (2, 1), facing each other across the line y = x
🔗 Inverse functions
①The inverse of y = aˣ is y = logₐ x
②If the point (p, q) is on one graph, the point (q, p) is on the other
③So the two graphs are symmetric about the line y = x
④The domain and range swap too: the logarithmic function takes all positive inputs and gives all real outputs
Properties of logarithmic functions
Definition
y = loga x (a > 0, a ≠ 1, x > 0)
Domain x > 0, range all real numbers
Two points every graph passes through
loga 1 = 0, loga a = 1
The graph passes through (1, 0) and (a, 1)
📐 Increasing, decreasing and the asymptote
①If a > 1, y grows as x grows: an increasing function
②If 0 < a < 1, y shrinks as x grows: a decreasing function
③The asymptote is the y-axis (x = 0)
④It is not defined for x ≤ 0
Logarithmic equations and inequalities
Log Equation
loga f(x) = loga g(x) ⟹ f(x) = g(x)
With the same base, compare the arguments, and check that they are positive
⚠️ Two things to watch in a log inequality
①If a > 1, logₐ M > logₐ N ⟺ M > N: the inequality keeps its direction
②If 0 < a < 1, logₐ M > logₐ N ⟺ M < N: the inequality flips
③At the end, combine the result with the conditions M > 0 and N > 0
Summary
Log Core
y = loga x: domain (0, ∞), range ℝ, asymptote x = 0
The inverse of the exponential function, its graph symmetric to the exponential graph about y = x
🎯 Exam Points
①The graph always passes through (1, 0), since logₐ 1 = 0
②Increasing if a > 1, decreasing if 0 < a < 1
③y = aˣ and y = logₐ x are inverse functions
④Log equations: match the bases, then compare the arguments
⑤Before giving an answer, check the condition argument > 0
Worked Examples & Past Exam
Example 1
Solve the logarithmic equation log2 (x − 1) = 3.
1
By the definition of a logarithm, the argument x − 1 is 2 cubed.
x - 1 = 23 = 8
2
x = 9, and the argument x − 1 = 8 is then positive, so it is the answer.
x = 9 (x - 1 = 8 > 0)
▸ x = 9
Rewriting a log equation in exponential form with the definition turns it into an ordinary equation. Just be sure to put the value back into the argument and check that it is positive.
Example 2
About which line are the graphs of y = log2 x and y = 2x symmetric?
1
Solving y = 2ˣ for x gives x = log₂ y, so the two functions are inverses of each other.
y = 2x ⟺ x = log2 y
2
Graphs of inverse functions pair up points with x and y swapped, so they are symmetric about the line y = x.
axis of symmetry: y = x
▸ Symmetric about the line y = x
If (1, 2) is on y = 2ˣ, then (2, 1) is on y = log₂ x. Swapping x and y is exactly reflecting in the line y = x.
2023 CSAT Math type, adapted
Solve the logarithmic inequality log3 (x − 2) < 2 (include the domain).
①2 < x < 11
②x < 11
③x > 2
④0 < x < 11
⑤2 < x < 9
▸ ① 2 < x < 11
1
The base 3 is greater than 1, so the inequality keeps its direction, and the condition x − 2 > 0 goes in too.
log3 (x-2) < 2 ⟺ 0 < x - 2 < 32
2
Adding 2 to every part of 0 < x − 2 < 9 gives 2 < x < 11.