Logarithmic Function Graph

How the Base Shapes a Logarithmic Graph

A logarithmic graph changes quickly near its vertical asymptote and then more slowly as its input grows. This shape appears whenever an output records the logarithm of a positive ratio, including sound-level and pH scales. The base determines whether the graph increases or decreases, while transformations move the graph and its asymptote.

Characteristics of Logarithmic Graphs

The following graph compares $y = \log_b x$ for several bases greater than $1$. Every graph passes through $(1,0)$ and increases at a different rate.

Properties of Logarithmic Graphs

For function $f(x) = \log_b x$ with $b > 1$:

1. Domain: $x > 0$ (positive numbers only)
2. Range: All real numbers ($-\infty < y < \infty$)
3. $x$-intercept: $(1, 0)$ because $\log_b 1 = 0$
4. Vertical asymptote: $y$-axis ($x = 0$)
5. Function behavior:
   • Increasing for $b > 1$
   • Decreasing for $0 < b < 1$

Drawing Logarithmic Function Graphs

The following concrete examples show each step for drawing a logarithmic function graph.

1. Drawing $y = \log_2 x$

   To draw this graph, we create a table of values by choosing $x$ values that are powers of $2$:

   $x$	$\frac{1}{8}$	$\frac{1}{4}$	$\frac{1}{2}$	$1$	$2$	$4$	$8$
   $y = \log_2 x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$

   Graph of $y = \log_2 x$

   Compare the intercept, asymptote, and direction of the curve.

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2. Drawing $y = \log_{\frac{1}{3}} x$

   For base $0 < b < 1$, the graph will be decreasing:

   $x$	$\frac{1}{27}$	$\frac{1}{9}$	$\frac{1}{3}$	$1$	$3$	$9$	$27$
   $y = \log_{\frac{1}{3}} x$	$3$	$2$	$1$	$0$	$-1$	$-2$	$-3$

   Graph of $y = \log_{\frac{1}{3}} x$

   Graph decreases because the base lies between $0$ and $1$.

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Comparing Logarithmic Graphs

Place logarithmic graphs with different bases on one coordinate system to compare their shapes directly:

Transformations of Logarithmic Graphs

A vertical shift moves the whole curve and leaves the vertical asymptote at $x = 0$. A horizontal shift instead moves the asymptote together with the $x$-intercept to the right or left, so the whole graph moves sideways.

Vertical Translation

We can shift the logarithmic function graph by adding or subtracting a constant $k$ to the function.

Horizontal Translation

We can shift the logarithmic function graph by adding or subtracting a constant $h$ to the function.

Exercises

The four problems below practice reading a logarithmic graph and predicting how a change inside the argument moves it. Work each one before the worked solutions, then compare your reasoning with the answers that follow.

1. Create a value table and draw the graphs of:

   • $y = \log_3 x$
   • $y = \log_{\frac{1}{2}} x$

2. Determine the domain, range, and asymptote of function $f(x) = \log_5 (x + 3)$.

3. If $f(x) = \log_2 x$ and $g(x) = \log_2 (x - 4)$, determine:

   • The shift of graph $g(x)$ relative to $f(x)$
   • The domain of $g(x)$

4. Sketch the graph of $y = \log_3 x + 2$ and determine the $y$-intercept.

Worked Solutions

1. Value tables:

   For $y = \log_3 x$:

   $x$	$\frac{1}{9}$	$\frac{1}{3}$	$1$	$3$	$9$
   $y$	$-2$	$-1$	$0$	$1$	$2$

   For $y = \log_{\frac{1}{2}} x$:

   $x$	$\frac{1}{4}$	$\frac{1}{2}$	$1$	$2$	$4$
   $y$	$2$	$1$	$0$	$-1$	$-2$

   Plot each ordered pair and draw a smooth curve that stays to the right of the vertical asymptote $x=0$. The base $3$ graph rises, while the base $\frac{1}{2}$ graph falls.

2. For $f(x) = \log_5 (x + 3)$:

   The logarithm is defined only when its argument is positive. The boundary of that condition gives the vertical asymptote:

   • Domain: $x + 3 > 0 \Rightarrow x > -3$
   • Range: All real numbers
   • Vertical asymptote: $x = -3$

3. For $g(x) = \log_2 (x - 4)$:

   Replacing $x$ with $x-4$ shifts the graph right. The same expression must also remain positive:

   • Shift: $4 \text{ units}$ to the right
   • Domain: $x - 4 > 0 \Rightarrow x > 4$

4. For $y = \log_3 x + 2$:

   Adding $2$ outside the logarithm shifts every point vertically without changing the domain or the asymptote:

   • Graph $y = \log_3 x$ shifted $2 \text{ units}$ up
   • There is no $y$-intercept because the domain is $x > 0$

   In particular, a $y$-intercept would require $x=0$, where the logarithm is undefined. The graph below confirms that it approaches but never reaches the $y$-axis.

   Sketch of Graph $y = \log_3 x + 2$

   Logarithmic graph base $3$ shifted $2 \text{ units}$ up.

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