Topics
6. Exponential & Logarithmic Functions
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United States of AmericaFL
Grade 912

6.04 Logarithmic functions

Lesson
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Lesson

Concept summary

We can use the inverse relationship between logarithmic and exponential functions to explore the graphs and characteristics of logarithmic functions, including natural logarithmic functions.

To draw the graph of a logarithmic function we can fill out a table of values for the function and draw the curve through the points found.

Because x^0=1 is true for any real number x, it can be rewritten as \log _x(1)=0, which tells us that any logarithm of 1 is 0, irrespective of the base used. This means the curve of any logarithmic function of the form y=\log _b\left(x\right) intersects the x-axis at (1,0). Similarly, as x^1=x is true for any real number x, so \log_b\left(b\right)=1 for any base b. This means that the curve of any logarithmic function of the form y=\log _b\left(x\right) will also pass through the point \left(b, 1\right).

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When the base b>1, the graph of the logarithmic function is a rising curve that increases at an increasing rate. As x approaches 0 the graph gets steeper but never meets the y-axis.

  • The graph is a strictly increasing function
  • The domain is \left(0, \infty\right)
  • The range is \left(-\infty, \infty\right)
  • The x-intercept is at \left(1,\, 0\right)
  • The base is 2
  • The vertical asymptote is x=0
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Where 0<b<1 the graph is a falling curve that decreases at a decreasing rate.

  • The graph is a strictly decreasing function
  • The domain is \left(0, \infty\right)
  • The range is \left(-\infty, \infty\right)
  • The x-intercept is at \left(1,\, 0\right)
  • The base is \dfrac{1}{2}
  • The vertical asymptote is x=0

Logarithmic functions can be dilated, reflected, and translated in a similar way to other functions.

The exponential parent function f\left(x\right)=\log_b\left(x\right) can be transformed to f\left(x\right)=a\log_b\left(x-h\right)+k

  • If a is negative, the basic curve is reflected across the x-axis
  • The graph is dilated vertically by a factor of a
  • The graph is translated to the right by h units
  • The graph is translated upwards by k units
  • The vertical asymptote, originally x=0, is translated horizontally to x=h
  • The domain of the graph becomes \left(h, \infty\right).

Worked examples

Example 1

Consider the values of the function f\left(x\right)=\log_{9}\left(x\right), and the transformed functions g\left(x\right) and h\left(x\right).

All values are rounded to two decimal places where appropriate.

x12349
f\left(x\right)00.320.50.631
g\left(x\right)-2-1.68-1.5-1.37-1
h\left(x\right)-0.32-0.5-0.63-0.73-1.05
a

Determine the transformations made to f\left(x\right) that would result in g\left(x\right).

Approach

We can see that the values of g\left(x\right) are all 2 less than the corresponding value for f\left(x\right).

Solution

Vertical translation of 2 units down.

b

Determine the transformations made to f\left(x\right) that would result in h\left(x\right).

Approach

We can see that h\left(1\right)=-f\left(2\right), h\left(2\right)=-f\left(3\right) and h\left(3\right)=-f\left(4\right).

Solution

Horizontal translation 1 unit left, and then reflected across the x-axis.

Example 2

Sketch the graph of the equation: y=3\log_2\left(x-3\right)+2

Approach

We can graph this equation by first completing a table of values and drawing the curve that passes through these points, or we can consider how the function f\left(x\right)=\log_2\left(x\right) has been transformed.

Considering each component separately, we can see that the graph of y=\log_2\left(x\right) is translated 3 units to the right, dilated vertically by a factor of 3, and then translated 2 units up.

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  • First the graph of f\left(x\right)=\log_2\left(x\right) is translated 3 units to the right.
  • This corresponds with the graph of \\y=\log_2\left(x-3\right).
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  • Next the graph of y=\log_2\left(x-3\right) is stretched vertically by a factor of 3 .
  • This corresponds with the graph of \\y=3\log_2\left(x-3\right).
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  • Finally the graph of y=3\log_2\left(x-3\right) is translated 2 units up.
  • This corresponds with the graph of \\y=3\log_2\left(x-3\right)+2.

Solution

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Reflection

We could also use a table of values to find some points and then draw the curve through these points. We know the asymptote is at x=3, so we want to start with a value slightly more than 3.

x3.013.13.54710
y-17.93-7.97-12810.42

We can see that these points correspond with points on our transformed graph.

Example 3

Consider the graph of the logarithmic function f\left(x\right).

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a

Sketch the inverse function on the same coordinate plane.

Approach

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  • We can graph the inverse relation by reflecting f\left(x\right) across the line y=x.
  • We can use the key points, that have integer values, at \left(-3, 0\right), \left(-2, 1\right), \left(0, 2\right), and \left(4, 3\right) and swap the x and y-values.
  • \left(-3, 0\right) \to \left(0, -3\right)
  • \left(-2, 1\right) \to \left(1, -2\right)
  • \left(0, 2\right) \to \left(2, 0\right)
  • \left(4, 3\right) \to \left(3, 4\right)
  • We can approximate a few more points on the inverse function and then sketch the curve through these inverse points.

The asymptote must also be reflected across the line y=x. This takes x=-4 to y=-4.

Solution

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b

Write the equation of the inverse function.

Approach

As f\left(x\right) is a logarithmic function, the inverse will be an exponential function of the form y=ab^{\left(x-h\right)}+k

We can see the asymptote is at y=-4, which indicates a vertical translation of 4 units down has occurred. This means the parent function would have a y-intercept of y=1, which indicates no stretch factor, and the initial value is a=1. It does not appear that there has been any horizontal shift, indicating h=0.

Solution

The inverse function is f^{-1}\left(x\right)=2^{x}-4.

Reflection

We can confirm this is the correct function by testing some of the other points:

2^1-4=-2, 2^2-4=0, and 2^3-4=4 as expected, based off the graph.

We could also determine the function f\left(x\right) to be f\left(x\right)=\log_2\left(x-4\right), by noticing that it is the graph of y=\log_2\left(x\right) translated left by 4 units, and then find its inverse algebraically.

Outcomes

MA.912.AR.5.8

Given a table, equation or written description of a logarithmic function, graph that function and determine its key features.

MA.912.AR.5.9

Solve and graph mathematical and real-world problems that are modeled with logarithmic functions. Interpret key features and determine constraints in terms of the context.

MA.912.F.1.1

Given an equation or graph that defines a function, determine the function type. Given an input-output table, determine a function type that could represent it.

MA.912.F.1.7

Compare key features of two functions each represented algebraically, graphically, in tables or written descriptions.

MA.912.F.3.7

Represent the inverse of a function algebraically, graphically or in a table. Use composition of functions to verify that one function is the inverse of the other.

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