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Logarithmic Function Graphs

Logarithmic Function Graphs

Logarithmic functions are the inverses of exponential functions. Understanding how to graph them and apply transformations is a fundamental skill in algebra and pre-calculus.

The Parent Graph: y=logb(x)y = \log_b(x)

The most basic logarithmic function is y=logb(x)y = \log_b(x), where the base b>0b > 0 and b1b \neq 1. No matter what base you use, the parent graph shares several key properties:

  • Domain: x>0x > 0. You can only take the logarithm of positive numbers.
  • Range: All real numbers (<y<-\infty < y < \infty).
  • x-intercept: The graph always passes through (1,0)(1, 0) because logb(1)=0\log_b(1) = 0.
  • Key Point: The graph passes through (b,1)(b, 1) because logb(b)=1\log_b(b) = 1.
  • Vertical Asymptote: The y-axis, or the line x=0x = 0. As xx gets closer to 00 from the right, the graph plunges downward toward -\infty (for b>1b > 1).

Comparing Different Bases

What happens when we change the base? Let's compare y=log2(x)y = \log_2(x) and y=log5(x)y = \log_5(x).

Both graphs pass through (1,0)(1, 0) and have a vertical asymptote at x=0x = 0. However, they differ in how steeply they curve:

  • For x>1x > 1, y=log5(x)y = \log_5(x) grows slower than y=log2(x)y = \log_2(x). For example, to reach a height of y=2y = 2, the base-2 graph only needs x=4x = 4, but the base-5 graph needs x=25x = 25.
  • For 0<x<10 < x < 1, y=log5(x)y = \log_5(x) is closer to the y-axis and x-axis than y=log2(x)y = \log_2(x).

Transformations of Logarithmic Graphs

You can move, stretch, and flip logarithmic graphs using the standard transformation formula:

y=alogb(xh)+ky = a\log_b(x - h) + k

  • hh (Horizontal Shift): Moves the graph left or right. This changes the domain and shifts the vertical asymptote to x=hx = h.
  • kk (Vertical Shift): Moves the graph up or down.
  • aa (Vertical Stretch/Compression): Stretches the graph vertically. If aa is negative, it reflects the graph across the x-axis.

Example: Graphing with Transformations

Problem: Graph y=log2(x3)+1y = \log_2(x - 3) + 1 and identify the domain and asymptote.

Solution:

  1. Identify the parent function: The parent function is y=log2(x)y = \log_2(x), which has a vertical asymptote at x=0x = 0 and key points at (1,0)(1,0) and (2,1)(2,1).
  2. Apply the horizontal shift: The (x3)(x - 3) inside the logarithm shifts the entire graph right by 3 units.
    • The new vertical asymptote is x=3x = 3.
    • The domain is found by setting the argument greater than zero: x3>0x - 3 > 0, so the domain is x>3x > 3.
  3. Apply the vertical shift: The +1+ 1 on the outside shifts the graph up by 1 unit.
  4. Find new key points:
    • Original (1,0)(1,0) \rightarrow shift right 3 to (4,0)(4,0) \rightarrow shift up 1 to (4,1)(4,1).
    • Original (2,1)(2,1) \rightarrow shift right 3 to (5,1)(5,1) \rightarrow shift up 1 to (5,2)(5,2).

To sketch the graph, draw a dashed vertical line at x=3x = 3 for the asymptote. Plot the points (4,1)(4,1) and (5,2)(5,2), and draw a smooth curve that passes through these points and approaches x=3x = 3 as it moves downward.