Section 4.4

Graphs of Logarithmic Functions

Log graphs appear on the final in domain, range, and transformation problems. Knowing how to find the vertical asymptote and domain also helps you spot extraneous solutions in section 4.6.

⚡ Quick Summary

If you take an exponential graph and flip it over the diagonal line $y = x$ (like folding a piece of paper along that line), you get a logarithmic graph. Everything swaps: the horizontal asymptote becomes a vertical asymptote, the point $(0, 1)$ becomes $(1, 0)$ , and the domain and range trade places. That's the whole connection — log graphs are just exponential graphs seen from a different angle.

Parent logarithmic function

f(x) = \log_b(x)

Passes through (1, 0); vertical asymptote at x = 0

Horizontal shift

f(x) = \log_b(x - h)

Asymptote moves to x = h; domain is (h, ∞)

📋 Before You Start

Tap any item if you need a refresher:

Plain-English version

The parent log curve $y = \log_b(x)$ hugs a vertical wall at $x = 0$ (it gets close but never touches), passes through the point $(1, 0)$ , and climbs slowly to the right. You can only plug in positive numbers — you can't take the log of zero or a negative number. That's why the domain is always "everything to the right of the wall."

Sliding the graph left or right moves the vertical wall with it. For example, $\log_b(x - 5)$ moves the wall from $x = 0$ to $x = 5$ , so now you can only plug in numbers bigger than 5. Sliding the graph up or down just raises or lowers the curve without moving the wall at all.

Here's the one-step trick for finding the domain of *any* log function: take whatever's inside the log and set it greater than zero. Solve that inequality, and you've got your domain. The boundary of that domain is also where the vertical asymptote lives. For example, if you have $\ln(3x - 6)$ , set $3x - 6 > 0$ to get $x > 2$ — that's the domain, and the asymptote is at $x = 2$ .

Key Formulas

Parent logarithmic function

f(x) = \log_b(x)

Passes through (1, 0); vertical asymptote at x = 0

Horizontal shift

f(x) = \log_b(x - h)

Asymptote moves to x = h; domain is (h, ∞)

Vertical shift

f(x) = \log_b(x) + k

Shifts graph up/down; asymptote and domain unchanged

General transformed log

f(x) = a \cdot \log_b(x - h) + k

a = vertical stretch/reflect, h = horizontal shift, k = vertical shift

Exponential-log reflection relationship

y = b^x \;\longleftrightarrow\; y = \log_b(x) \text{ reflected over } y = x

Key Takeaways

✓Log graphs are the reflection of exponential graphs over the line $y = x$ — swap the roles of $x$ and $y$
✓The parent $y = \log_b(x)$ has a vertical asymptote at $x = 0$ and passes through $(1, 0)$
✓Horizontal shifts move the vertical asymptote: $\log_b(x - h)$ has VA at $x = h$ and domain $(h, \infty)$
✓To find the domain of any log function, set the argument $> 0$ and solve

⚠️ Common Mistakes

✗Confusing which function has which asymptote: exponentials have horizontal asymptotes; logs have vertical asymptotes
✗Thinking a vertical shift ( $+k$ ) moves the vertical asymptote — only horizontal shifts change where the asymptote is
✗Forgetting to flip the inequality when dividing by a negative while finding the domain (e.g., $-2x > -6$ becomes $x < 3$ )
✗Setting the argument $\geq 0$ instead of $> 0$ — you cannot take the log of zero, only positive numbers

Next: Section 4.5 Builds on: 4.3 Builds on: 4.2

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4.3 Logarithmic Functions

4.5 Logarithmic Properties