Logarithmic Graphs

Video lesson · from 1:34:132 worked examples

One lesson video covers all of Functions & Graphs, so it opens at 1:34:13 for this topic — not from the beginning.

Theory

The Logarithmic Graph

Since the logarithmic function is the inverse of the exponential function, we can reflect the exponential graph in the line $y = x$ to find the logarithmic graph.

When $x = 1, f(1) = \log_a 1 = 0 \implies (1, 0)$
When $x = a, f(a) = \log_a a = 1 \implies (a, 1)$

Every logarithmic graph of the form $f(x) = \log_a x$ passes through the points $(1, 0)$ and $(a, 1)$ .

⚠️ Common Examiner Traps

Every log graph passes through $(1,0)$ : because $\log_a 1 = 0$ . This is the counterpart of $(0,1)$ on the exponential.
The $y$ -axis is the asymptote: vertical, not horizontal — the opposite of the exponential graph. There is no $y$ -intercept.
Only defined for $x \gt 0$ : nothing at all is drawn to the left of the $y$ -axis.
Transformations shift the asymptote: $\log_a(x-2)$ moves the vertical asymptote to $x = 2$ , and the graph with it.

Worked examples

Example 1

Transformations

Sketch the graph of $y = \log_5(x - 3)$ .

Base graph is $y = \log_5 x$ passing through $(1,0)$ and $(5,1)$ with asymptote $x=0$ .

The $-3$ inside shifts the graph RIGHT by 3 units.

$(1,0) \to (4,0)$
$(5,1) \to (8,1)$
Asymptote moves to $x = 3$

Example 2

Missing Constants

Part of the graph of $y = \log_m(x + k)$ is shown.

It passes through $(-4, 0)$ and $(5, 1)$ .

Find $k$ and $m$ .

Substitute point $(-4, 0)$ :

0 = \log_m(-4 + k)

Convert to exponential form:

m^0 = -4 + k \implies 1 = -4 + k \implies k = 5

Now equation is $y = \log_m(x + 5)$ . Substitute point $(5, 1)$ :

1 = \log_m(5 + 5) \implies 1 = \log_m(10)

m^1 = 10 \implies m = 10

So $k = 5$ and $m = 10$ .