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3. Exponential & Logarithmic Functions
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United States of AmericaTennessee
Integrated Math 3

3.03 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.

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The graph of f\left(x\right)=b^x has a point at \left(0,1\right), a point at \left(1,b\right), and an asymptote at y=0.

As a table:

Since f\left(x\right)=\log_b(x) is the inverse of f\left(x\right)=b^x its graph has a point at \left(1,0\right) and \left(b,0\right) and an asymptote of x=0.

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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

Determine the equation of the asymptote and two points on the curve.

Approach

The asymptote is displayed visually as a dashed line that the function approaches, but does not cross.

Solution

The equation of the asymptote is x=-4 and the graph has points at \left(-3,0\right) and \left(-2,1\right).

Reflection

There are also points located at \left(0,2\right) and \left(4,3\right).

b

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)

This allows us to approximate a few more points on the inverse function and then sketch the curve through these inverted points.

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

Solution

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c

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.

By analyzing the change in y-values we can determine the value of the base, b.

Since the ratio of the differences is 2, the base of the exponential function is 2.

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

M3.N.Q.A.1

Use units as a way to understand real-world problems.*

M3.F.IF.B.3

For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.*

M3.F.IF.B.4

Calculate and interpret the average rate of change of a function (presented algebraically or as a table) over a specified interval. Estimate and interpret the rate of change from a graph. *

M3.F.IF.C.5

Graph functions expressed algebraically and show key features of the graph by hand and using technology.*

M3.F.IF.C.6

Compare properties of functions represented algebraically, graphically, numerically in tables, or by verbal descriptions.*

M3.F.IF.C.6.A

Compare properties of two different functions. Functions may be of different types and/or represented in different ways.

M3.F.IF.C.6.B

Compare properties of the same function on two different intervals or represented in two different ways.

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP2

Reason abstractly and quantitatively.

M3.MP3

Construct viable arguments and critique the reasoning of others.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

M3.MP7

Look for and make use of structure.

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