Practice Maths

Graphs of Logarithmic Functions

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Key Terms

Basic shape
y = loga(x) has domain x > 0, range all reals, vertical asymptote x = 0, x-intercept (1, 0).
If a > 1
function is increasing (e.g. y = log2(x)).
If 0 < a < 1
function is decreasing (e.g. y = log0.5(x)).
General form
y = A logb(x − h) + k
h: horizontal shift right (domain x > h, VA: x = h)
k: vertical shift up
A: vertical dilation (stretch if |A|>1, compression if |A|<1; reflection if A<0)
x-intercept
set y = 0 and solve logb(x − h) = −k/A.
📚 Key Features: y = A logb(x − h) + k
Domain: x > h    Vertical Asymptote: x = h
Range: all real numbers
x-intercept: solve A logb(x − h) + k = 0
y-intercept: substitute x = 0 (only exists if h < 0)
Function Asymptote x-intercept Effect on y=log2(x)
y = log2(x)x = 0(1, 0)Base function
y = log2(x − 3)x = 3(4, 0)Shift right 3
y = log2(x) + 2x = 0(1/4, 0)Shift up 2
y = 2log2(x)x = 0(1, 0)Vertical stretch ×2
y = −log2(x)x = 0(1, 0)Reflect in x-axis
x y x=0 ½ 1 2 4 −1 0 1 2 (½, −1) (1, 0) (2, 1) (4, 2) y = log₂(x)
Hot Tip The x-intercept of y = loga(x − h) + k is found by setting y = 0: loga(x − h) = −k, so x − h = a−k, giving x = h + a−k. This is a key exam technique.

Worked Example 1 — Key features of a transformed log

Find the domain, asymptote, x-intercept, and y-intercept of y = log2(x − 1) + 3.

Domain: x − 1 > 0 → x > 1

Vertical asymptote: x = 1

x-intercept (y = 0): log2(x − 1) = −3 → x − 1 = 2−3 = 1/8 → x = 1 + 1/8 = 9/8

y-intercept: x = 0 is not in the domain (x > 1), so no y-intercept

Worked Example 2 — Finding the equation from a graph

A log function has vertical asymptote x = 2, passes through (3, 0) and (10, 1). Find its equation.

Form: y = A log10(x − 2) + k (base 10, VA at x=2)

At (3, 0): 0 = A log(1) + k = A(0) + k → k = 0

At (10, 1): 1 = A log(8) → A = 1/log(8) ≈ 1/0.903 ≈ 1.107

Equation: y = (1/log 8) log(x − 2)   or equivalently   y = log8(x − 2)

Introduction

The graph of a logarithmic function is the mirror image of an exponential graph, reflected in the line y = x. This makes sense because logarithms and exponentials are inverses. Understanding the shape, key features, and how transformations move or stretch the curve is essential for both graphing questions and modelling.

The Basic Graph: y = loga(x)

For any base a > 1, the graph of y = loga(x) has these fixed features regardless of the exact base:

  • Domain: x > 0 — you cannot take the log of a non-positive number
  • Range: all real numbers (−∞, ∞)
  • Vertical asymptote: x = 0 (the y-axis) — the curve approaches but never touches
  • x-intercept: (1, 0) — because loga(1) = 0 for any base
  • Passes through (a, 1) — because loga(a) = 1
  • Increasing for a > 1; decreasing for 0 < a < 1

Transformations of y = loga(x)

The general transformed form is y = A logb(x − h) + k. Each parameter changes one feature of the graph:

Horizontal shift (h): y = logb(x − h)

Moves the entire graph right by h units. The vertical asymptote moves from x = 0 to x = h. The domain becomes x > h. The x-intercept shifts from (1, 0) to (1 + h, 0).

Example: y = log2(x − 3): asymptote x = 3, domain x > 3, x-intercept at (4, 0) since log2(1) = 0.

Vertical shift (k): y = logb(x) + k

Moves the entire graph up by k units. The asymptote stays at x = 0 (domain unchanged). The x-intercept moves: set logb(x) + k = 0, so logb(x) = −k, giving x = b−k.

Example: y = log2(x) + 2: x-intercept where log2(x) = −2 → x = 1/4. So new x-intercept is (1/4, 0).

Vertical dilation (A): y = A logb(x)

Stretches the graph vertically. A > 1 makes the curve steeper; 0 < A < 1 flattens it; A < 0 reflects it in the x-axis (turns it upside down). The x-intercept (1, 0) is unchanged (since A × 0 = 0). A reflected graph is decreasing even if b > 1.

Sketching a Transformed Log Graph: 5-Step Method

  1. Identify h: draw the vertical asymptote x = h as a dashed line.
  2. Find the domain: x > h.
  3. Find the x-intercept: solve A logb(x − h) + k = 0.
  4. Find a second point: substitute a convenient x value (e.g. x = h + b to get y = A + k).
  5. Sketch the curve approaching the asymptote, passing through your points, and curving in the correct direction.
Worked Example 1 — Sketch y = 2log3(x + 1) − 2

Step 1: h = −1 → VA: x = −1 (dashed). Domain: x > −1.

Step 2: x-intercept: 2log3(x+1) = 2 → log3(x+1) = 1 → x+1 = 3 → x = 2. Point: (2, 0).

Step 3: y-intercept: x=0: 2log3(1)−2 = 2(0)−2 = −2. Point: (0, −2).

Step 4: At x = 8: 2log3(9)−2 = 2(2)−2 = 2. Point: (8, 2).

Step 5: Sketch: increasing curve, approaches x=−1 from the right, passes through (0,−2), (2,0), (8,2).

💡 Key Reminder: The range of a log function is always all real numbers, regardless of any vertical transformations. Only the domain and asymptote change with horizontal transformations. Always state the domain explicitly in any graphing question.

Summary

y = A logb(x − h) + k: VA at x = h, domain x > h, range all reals, x-intercept found by setting y = 0. Horizontal shift changes the asymptote; vertical shift changes the x-intercept; vertical dilation stretches/reflects the graph.

Mastery Practice

See Answers ➔
  1. Fluency

    State the domain, range, and vertical asymptote of each function.

    1. (a) y = log2(x)
    2. (b) y = log3(x − 4)
    3. (c) y = log(x + 2)
    4. (d) y = ln(x − 5)
  2. Fluency

    Find the x-intercept of each function.

    1. (a) y = log2(x) − 3
    2. (b) y = log(x + 5)
    3. (c) y = 2log3(x − 2) + 4
    4. (d) y = log5(x) + 1
  3. Fluency

    Describe the transformation(s) applied to y = log2(x) to obtain each function.

    1. (a) y = log2(x) + 5
    2. (b) y = log2(x − 3)
    3. (c) y = 3log2(x)
    4. (d) y = −log2(x)
  4. Fluency

    Find the y-intercept of each function (if it exists).

    1. (a) y = log3(x + 3)
    2. (b) y = log2(x − 1) + 2
    3. (c) y = log(2x)
    4. (d) y = ln(x + e)
  5. Understanding

    Sketch each function. Label the vertical asymptote, x-intercept, and one other point.

    Method: Find VA, domain, x-intercept, y-intercept (if exists), and one convenient extra point. Then sketch.
    1. (a) y = log2(x + 4)
    2. (b) y = log3(x) − 1
  6. Understanding

    For each function, state whether it is increasing or decreasing, find the domain, and find the x-intercept.

    1. (a) y = log0.5(x)
    2. (b) y = −2log3(x + 1)
    3. (c) y = log4(x − 3) + 2
  7. Understanding

    Write the equation of a log function (base 2) that has been:

    1. (a) shifted right 5 units and up 3 units
    2. (b) vertically stretched by a factor of 4 and reflected in the x-axis
    3. (c) shifted left 1 unit, vertically compressed by factor 1/2, and shifted down 2 units
  8. Understanding

    Find the equation of the logarithmic function.

    1. (a) The function has vertical asymptote x = 1, x-intercept (4, 0), and passes through (10, 1). Write in the form y = logb(x − 1).
    2. (b) The function has the form y = A log(x) + k, passes through (1, −3) and (10, 2). Find A and k.
  9. Problem Solving

    Comparing log graphs.

    Challenge. Two functions are f(x) = log2(x) and g(x) = log2(4x) − 1.
    1. (a) Show that f(x) = g(x) for all x in the domain.
    2. (b) Hence explain why the two graphs are identical.
  10. Problem Solving

    A log model for loudness.

    Challenge. The loudness L (in decibels) perceived by the human ear is modelled by L = 10log(I/I0), where I is the intensity and I0 = 10−12 W/m2.
    1. (a) Write L as a function of I without I0 in the expression.
    2. (b) State the domain of this function in terms of I.
    3. (c) What value of I gives L = 0? Interpret your answer.
    4. (d) Sketch L vs I for 10−12 ≤ I ≤ 1, labelling two key points.