# Logarithmic Graphs

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The graph of $y = \log_a(x)$ (where a is positive but excluding 1) is called a **logarithmic graph**.

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Logarithmic graphs (base 10) are used in measuring

- levels of noise (decibels)
- strength of earthquakes (Richter Scale)

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The graph of $y = \log_a(x)$

- has a vertical asymptote at $x = 0$
- is only defined for $x > 0$
- has an x-intercept at $(1, \; 0)$
- goes through the point $(a,\; 1)$

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## Graph of Log Base 2

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The graph of $y = \log_2(x)$ is shown here:

- Asymptote: $x = 0$

- x-intercept: $(1, \; 0)$ because $\log_2(1) = 0$

- A second point is at $(2, \; 1)$ because $\log_2(2) = 1$

- No y-intercepts

- No turning points

- Domain: $x \in R^+$

- Range: $y \in R$

- This is a strictly increasing graph (gradient is always positive).

- This graph is one-to-one

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## Comparing Log and Exponential Graphs

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The graphs of $y = a^x \text{ and } y = \log_a(x)$ are the **inverse** of each other.

- This means that they are a reflection across the line $y = x$.

- the
**x**and**y**coordinates of the individual points are swapped

- so $(x, \; y) \rightarrow (y, \; x)$

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## Log Graphs with a > 1

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All logarithmic graphs, $y = \log_a(x)$, where **a > 1** have the following in common:

- Asymptote at $x = 0$:
- as $x \rightarrow 0, \; y \rightarrow -\infty$

- x-intercept at $(1, \; 0): \; \log_a(1) = 0$

- 2nd point at $(a, \; 1): \; \log_a(a) = 1$

- Strictly Increasing graph
- As $x \rightarrow +\infty, \; y \rightarrow +\infty$

- Domain: $x \in R^+$

- Range: $y \in R$

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## Log Graphs with 0 < a < 1

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All logarithmic graphs, $y = \log_a(x)$, where **0 < a < 1** have the following in common:

- Asymptote at $x = 0$:
- as $x \rightarrow 0, \; y \rightarrow +\infty$

- x-intercept at $(1, \; 0):\; \log_a(1) = 0$

- 2nd point at $(a, \; 1): \; \log_a(a) = 1$

- Strictly Decreasing Graph
- As $x \rightarrow +\infty, \; y \rightarrow -\infty$

- Domain: $x \in R^+$

- Range: $y \in R$

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**Note**

… … $\log_{0.5}(x) = -\log_2(x)$

… … so using the transformation rules, we can see that $y = \log_{0.5}(x)$

… … is the reflection of $y = \log_2(x)$ across the x-axis (in the y-direction)

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## Log Graphs on the Classpad

- Most scientific calculators have two bases of logarithm built into them.
- $\log_{10}(x)$ … … (usually represented as
**LOG**) - $\log_e(x)$ … … (usually represented as
**LN**)

- $\log_{10}(x)$ … … (usually represented as

- The classpad has a function to work in any base so it is easy to draw log graphs with any base.

- If you find yourself using a graphing calculator that can only use the two "standard" bases,
- You may need to use the change of base log law.

- eg, to draw the graph of $y = \log_2(x)$,

- you would have to enter $y = \dfrac{\log_{10}(x)}{\log_{10}(2)}$

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## Transformations

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We can use the standard transformations such as dilations, reflections and translations on the $y = \log_a(x)$ graph.

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#### Example 2

Sketch the graph of $y = \log_3(2x - 3)$

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**Method 1**

… … Asymptote occurs at $y = \log_3(0) \quad \log_3(0)$ is undefined

… … $2x - 3 = 0$

… … $2x = 3$

… … $x = \frac{3}{2}$ … … equation of asymptote

… … x-intercept occurs at $y = \log_3(1) \quad \{ \log_3(1) = 0 \}$

… … $2x - 3 = 1$

… … $2x = 4$

… … $x = 2$ … … {x-intercept: $(2, \; 0)$}

… … A 2nd point occurs at $y = \log_3(3) \quad \{ \log_3(3) = 1 \}$

… … $2x - 3 = 3$

… … $2x = 6$

… … $x = 3$ … … {2nd point: $(3, \; 1)$ }

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**Method 2**

… … $y = \log_3(2x - 3)$

… … can be written as:

… … $y = \log_3 \big( 2(x - \frac{3}{2} ) \big)$

… … This is the $y = \log_3(x)$ graph with transformations:

… … … dilated by a factor of $\frac{1}{2}$ from the y-axis (in the x-direction)

… … … translated by $\frac{3}{2}$ units parallel to the x-axis (to the right)

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… … In matrix form, this is:

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{2} & 0 \\ 0 & 1 \\ \end{matrix} \right] \; \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} \frac{3}{2} \\ 0 \\ \end{matrix} \right]$

… … OR

… … $x'=\frac{1}{2}x+\frac{3}{2}$

… … $y' = y$

… … Asymptote: $x = 0 \rightarrow x' = \frac{3}{2}$

… … x-intercept: $(x = 1, \; y = 0) \rightarrow (x' = 2, \; y' = 0)$

… … 2nd point: $(x = 3, \; y = 1) \rightarrow (x' = 3, \; y' = 1)$

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