6.03 Logarithmic functions

Identify key characteristics of a logarithmic graph, such as the approach to a vertical asymptote and increasing or decreasing behavior in y-values as x increases.

The graph shows a curve that increases as x gets larger and has a vertical asymptote on the left, which are typical traits of a logarithmic function.

This graph represents a logarithmic function.

This graph does not match the typical shape of a logarithmic function. It shows a graph with two asymptotes and a discontinuity that divides the function into two sections, which is characteristic of a rational function instead.

So this graph does not represent a logarithmic function.

Analyze the given table of values to see if it corresponds to a logarithmic function. We can use the relationship between exponential functions and logarithmic functions to identify patterns.

The y-values in the table increase by 1 each time the x-value doubles, which suggests a logarithmic relationship where the base of the logarithm is 2. We can also see that this table is a possible inverse of the function y=2^x, which further verifies that the relationship is logarithmic.

This table represents a logarithmic function.

Use the graph of f\left(x\right)=\left(\dfrac{1}{2}\right)^x to create a table of values for the logarithmic function.

Identify points on the graph of f\left(x\right) to create a table of values. Each point (x,\,y) on f\left(x\right) will correspond to the point (y,x) on the inverse, g\left(x\right).

To create a table of values, we will look for points on f\left(x\right) that are easy to identify.

Now, we will interchange the inputs and outputs to create a table of values for the inverse.

Graph g\left(x\right)=\log_\frac{1}{2}\left(x\right).

Plot the points from the table created in part (a) and draw the curve that passes through these points.

Compare the asymptotes, domain, range, and intercepts of f\left(x\right)=\left(\dfrac{1}{2}\right)^x and \\g\left(x\right)=\log_\frac{1}{2}\left(x\right).

Since these are inverse functions, the features related to inputs and outputs will be swapped. We can use this understanding and our knowledge of parent exponential and logarithmic functions to compare their key features.

• Asymptotes: For f\left(x\right), there is a horizontal asymptote at y=0, and for g\left(x\right), there is a vertical asymptote at x=0.

• Domain: The domain of f\left(x\right) is all real numbers, while the domain of g\left(x\right) is x > 0.

• Range: The range of f\left(x\right) is y > 0, and the range of g\left(x\right) is all real numbers.

• Intercepts: For f\left(x\right), the y-intercept is at (0,1), and there are no x-intercepts. For g\left(x\right), the x-intercept is at (1,0), and there is no y-intercept.

Notice that both functions are decreasing over their domain. For exponential functions of the form y=b^x and logarithmic functions of the form y=\log_b{x}, the functions will increase when b>1 and decrease when 0<b<1.

Identify the increasing or decreasing intervals.

The base of a parent logarithmic function will tell us whether the function is increasing or decreasing over its domain.

• If b>1, the function increases over the domain \left(0,\infty\right)

• If 0<b<1, the function decreases over the domain \left(0,\infty\right)

The base of the logarithm is 3. Since b>1, this tells us it is an increasing function.

The function is increasing for (0, \infty)

We can use the graph of y=3^x to find the key points and use them to graph f\left(x\right)=\log_3\left(x\right)

Graph of y=3^x

Graph of f\left(x\right)=\log_3\left(x\right)

We can confirm from the graph that the function f\left(x\right)=\log_3\left(x\right) is increasing everywhere.

Determine the end behavior.

To determine the end behavior, we have to look at what is happening with our graph as the x-values decrease and increase. Remember that a graph can never cross a vertical asymptote, so this will impact the values x can approach as they decrease. Instead of approaching negative infinity, x now approaches a single value.

We can use the graph from the reflect and check of part (a) to visualize the end behavior.

Now let's take a look at the right side of the graph. As the x-values increase, f\left(x\right) increases infinitely. As x\to \infty, f\left(x\right)\to \infty.

State the domain and range of f\left(x\right)=\log_3\left(x\right).

A function's domain is the set of all possible inputs, and a function's range is the set of all possible outputs.

Domain: (0, \infty)

Range: (-\infty,\,\infty)

Remember, the graph never reaches an asymptote, but rather continues approaching it inifinitely. This graph has a vertical asymptote at x=0, which means this graph will never reach x=0. This is why we did not include x=0 in the domain.

Evaluate f\left(x\right)=\log_3\left(x\right) when x=81.

Begin by substituting x=81 into the function. Then ask yourself: to what exponent must the base be raised to in order to get 81?

This question is asking us to find the output of the expression \log_3(81).

The base of the logarithm is 3. To what exponent must 3 be raised to get 81? We must raise 3 to the power of 4 to get 81.

Because 3^4=81,\,\log_3(81)=4.

We can also use technology to check our answer. However, many scientific calculators will only evaluate logarithms with a base of 10, so we must be careful to use one that allows us to specify the base of the logarithm.

Parent function: f\left(x\right) = \log_{\frac{1}{3}}x

To write the equation of the transformed function, identify the transformations applied to the parent function. Look for reflections, translations, stretches, or compressions.

The transformed graph represents a reflection of the parent function over the y-axis. This can be represented by negating the input variable x.

The transformed function is f\left(x\right) = \log_{\frac{1}{3}}\left(-x\right).

Parent function: f\left(x\right) = \log_{4}x

The shape of the transformed graph appears to be the same as the original graph. This implies that the function has not been dilated.

The transformed graph shows the parent function was shifted upwards by 2 units. This vertical translation can be represented by adding 2 to the parent function.

The transformed function is f\left(x\right) = \log_{4}\left(x\right) + 2.

Describe how g\left(x\right)=\log(-(x+2))-1 was transformed from f\left(x\right).

To describe the transformation, compare g\left(x\right) to the transformation form of a function: g\left(x\right)=a\log\left[c\left(x-h\right)\right]+k

When comparing g\left(x\right) to transformation form, we can see that a=1,\,c=-1,\,h=-2,\, and k=-1.

• c=-1 indicates a reflection over the y-axis
• h=-2 indicates a horizontal shift 2 units to the left
• k=-1 indicates a vertical shift 1 unit downward

Combining these transformations, g\left(x\right) is the result of reflecting f\left(x\right) across the y-axis, shifting it 2 unit to the left and 1 unit down.

Graph g\left(x\right) on the same coordinate plane.

Apply the identified transformations to the original graph point by point, reflecting across the y-axis, shifting left 2 units and down 1 unit to graph g\left(x\right).

We can use the points \left(1,\,0\right) and \left(10,\,1\right) on the graph of f\left(x\right) to graph the transformed function.

We can use technology to confirm that we graphed the function correctly and to verify the transformations identified in part (a).

This graph has the same points as the function we plotted, showing that we graphed the function correctly. We can also see that the function has been reflected across the y-axis and shifted 2 units left and 1 unit down.

Identify and compare the asymptotes, zeros, and end behavior of f\left(x\right) and g\left(x\right).

To compare these functions, we'll first identify each feature for both functions separately. We will look for the vertical asymptote by finding the value that makes the argument of the logarithm zero, and we can find the zeros and determine the end behavior by examining the graph from part (b).

For f\left(x\right)=\log\left(x\right), the vertical asymptote is at x=0, since the logarithm is undefined at x=0.

For g\left(x\right)=\log\left(-\left(x+2\right)\right)-1, we know that the vertical asymptote will be shifted left 2 units due to the horizontal translation. This function has a vertical asymptote at x=-2 as this logarithm is undefined at x=-2.

Now, let's use the graph from the previous part to find and compare the zeros of each function:

When examining the end behavior, we must take the vertical asymptote into consideration. Looking at the graph of f\left(x\right), we see as x approaches positive infinity, the y-valuesapproach positive infinity. As the x-valuesdecrease, the function approaches the vertical asymptote at {x=0} from the positive (right) side, and the y-values approach negative infinity.

• End behavior as x\to 0^{+}, f\left(x\right)\to -\infty

• End behavior as x\to \infty, f\left(x\right)\to \infty

Now let's take a look at the end behavior of g\left(x\right). As the x-valuesapproach negative infinity, the y-values approach positive infinity. As the x-values increase, the function approaches the vertical asymptote at x=-2 from the negative (left) side, and the y-values approach negative infinity.

• End behavior as x\to -\infty, f\left(x\right)\to \infty

• End behavior as x\to -2^{-}, f\left(x\right)\to -\infty

Upon comparison, the end behavior indicates f\left(x\right) increases everywhere whereas g\left(x\right) decreases everywhere.

We can determine the zeros of logarithmic functions algebraically by setting the function equal to 0 and solving for x.

To find the zero of f\left(x\right), we let \log\left(x\right)=0. In a logarithmic function, x is the result of raising the base to the power of 0. Any value raised to the 0th power is 1, so x=1.

Similarly, we let g\left(x\right)=0 to find the zero of g\left(x\right). We must first isolate the logarithmic expression to determine the exponent that the base must be raised to.

Now, we know that the base (which in this case is 10) raised to the power of 1 must be equal to the argument of the logarithm.