Logarithmic Functions - Logarithms - High School Algebra II Unlocked (2016)

High School Algebra II Unlocked (2016)

Chapter 5. Logarithms

Lesson 5.4. Logarithmic Functions

REVIEW

For a function f(x):

• The graph of f(−x) is the reflection of f(x) across the y-axis.

• The graph of −f(x) is the reflection of f(x) across the x-axis.

• Multiplying the x-value or the function value by a constant changes the scale of f(x).

In Lesson 5.1, we learned that functions f(x) and g(x) are inverses of one another if f(g(x)) = x and g(f(x)) = x. In Lesson 5.2, we learned that logbbx = x and blogbx = x. If we define f(x) = bx and g(x) = logbx, then this demonstrates that f(x) and g(x) are inverse functions.

f(g(x)) = f(logbx) = blogbx = x

g(f(x)) = g(bx) = logbbx = x

So, f(x) = bx and g(x) = logbx are inverse functions, reflecting how exponentiation and logarithm are inverse operations. The inverse functions for more complicated exponential functions are also logarithmic functions.

Given that f(x) = 5 ⋅ 72x, what is f −1(x)?

Replace f(x) with y, switch the x and y variables, and solve for the new y.

y = 5 ⋅ 72x

x = 5 ⋅ 72y

Switch x and y.

x/5 = 72y

Divide both sides by 5.

log7(x/5) = 2y

Take the logarithm, base 7, of both sides.

log7x − log75 = 2y

Use the logarithmic quotient property.

= y

Divide both sides by 2.

So, f −1(x) = .

What is the inverse function of p(x) = 2 ⋅ 4x + 9?

Use y for p(x), switch the x and y variables, and solve for the new y.

y = 2 ⋅ 4x + 9

x = 2 ⋅ 4y + 9

Switch x and y.

x/2 = 4y + 9

Divide both sides by 2.

log4(x/2) = y + 9

Take the logarithm, base 4, of both sides.

log4x − log42 = y + 9

Use the logarithmic quotient property.

log4x − 1/2 = y + 9

Evaluate log42.

log4x − 19/2 = y

Subtract 9 from both sides.

So, p−1(x) = log4x − 19/2.

Alternatively, after switching x and y, we could have rewritten the right side of the equation to be a single exponential expression of base 2 raised to a power.

x = 2 ⋅ (22)y + 9

Rewrite 4 as 22.

x = 2 ⋅ 22y + 18

Raise 22 to the power of (y + 9) by multiplying the exponents.

x = 22y + 19

Combine the two base-2 exponential terms: 21 ⋅ 22y + 18 = 21 + 2y + 18.

log2x = 2y + 19

Take the logarithm, base 2, of both sides.

log2x − 19 = 2y

Subtract 19 from both sides.

Divide both sides by 2.

This answer (p−1(x) = or p−1(x) = − 19/2) would be fine for the given question, but it doesn’t look the same as our first answer. Let’s use the properties of logarithms to show that they are actually equivalent. We want to convert this equation to our original answer, y = log4x − 19/2, which uses a base-4 logarithm.

y = − 19/2

y = − 19/2

Use the change of base property to write log2x as a quotient of base-4 logarithms.

y = − 19/2

Evaluate log42.

y = − 19/2

Replace division of log4x by 1/2 with multiplication by its reciprocal, 2.

y = log4x − 19/2

Cancel out the common factor of 2.

This produces the same inverse function equation that we originally found, p−1(x) = log4x − 19/2.

GRAPHING LOGARITHMIC FUNCTIONS

Logarithmic functions are inverses of exponential functions, so their graphs are related.

Find the inverse function of f(x) = 2x and graph both functions on the same coordinate grid. Compare the two graphs.

y = 2x

Replace f(x) with y.

x = 2y

Switch x and y.

log2x = y

Take the logarithm, base 2, of both sides.

As we learned in
Lesson 5.1, the graphs
of a function and its
inverse function are
reflections across
the line y = x. This
can help you graph
logarithmic functions,
as reflections of
exponential functions.

You may also recognize
that this is an example
of an exponential
function of the form
f(x) = bx, which has
an inverse function of
f −1(x) = logbx, with a
b-value of 2. So, the
function f(x) = 2x has
an inverse function
of f −1(x) = log2x.

So, f −1(x) = log2x.

Using graphing technology or tables of values, graph f(x) and f −1(x) on the same coordinate grid.

Switching x and y in
the given exponential
equation replaced each
point (a, b) with point
(b, a), reflecting the
graph of f(x) across
the line y = x, as we
also saw previously
in Example 2 for a
quadratic function.

These inverse functions are reflections of one another across the line y = x. Both graphs are always increasing. The domain of f(x) = 2x is all real numbers, while its range is all real numbers greater than 0 (there is a horizontal asymptote at the x-axis for its left arm). The domain of f −1(x) is all real numbers greater than 0 (there is a vertical asymptote at the y-axis for its left arm), while its range is all real numbers. This illustrates how the domain of a function is the range of its inverse and the range of a function is the domain of its inverse.

A logarithmic function of the form y = logbx has a domain of x > 0 and a range of all real numbers.

Recall from Lesson 5.2
that logb0 and logbc for
c < 0 are undefined. In
other words, y = logbx
is only defined for
x-values greater than
0, which is apparent
from its graph.

In Lesson 5.2, you
learned that
logb1 = 0 and
logbb = 1. Because
logb1has a value of 0,
no matter what value
b has, a function of the
form y = logbx always
has an x-intercept
of 1. When x = b in a
function of this form,
y = logbb, which
equals 1. So,
y = logbx passes
through the
point (b, 1).

The graph of f(x) = 2x passes through (0, 1), because 20 = 1, while f −1(x) = log2x passes through (1, 0), because log21 = 0. The graph f(x) = 2x also passes through (1, 2), while f −1(x) = log2x passes through (2, 1). This illustrates how the inverse of a function transforms each (x, y) point on the original function to (y, x).

A logarithmic function of the form y = logbx has an x-intercept of 1, meaning that it passes through the point (1, 0). A function y = logbx also passes through the point (b, 1).

Notice also that the graph of f −1(x) = log2x is below the x-axis for positive values less than 1. This reflects the fact that a base-2 logarithm of a proper fraction must be negative. Remember the meaning of logarithm as an exponent. The base 2 must be raised to a negative exponent to produce an x-value of 1/2, 1/4, or any positive value less than 1.

Compare the graphs of f(x) = log3x and g(x) = log1/3x.

We know that the graph of f(x) = log3x passes through (1, 0) and (3, 1), and it has a vertical asymptote of the y-axis for its left end.

The graph of g(x) = log1/3x also has a vertical asymptote of the y-axis for its left end, and passes through the points (1, 0) and (1/3, 1). When x = 3, log1/3x = −1, so the graph also passes through (3, −1). The graph of g(x) = log1/3x is shown below.

As you can see, this graph is always decreasing, whereas f(x) was always increasing. In fact, the graph of g(x) is the reflection of f(x) across the x-axis. If you rewrite the relationship g = log1/3x as (1/3)g = x, it is the same as 3g = x, or log3x = −g. So, g = −log3x. In other words, g(x) = −f(x), so the graph of g(x) is a reflection of f(x) across the x-axis.

A logarithmic function of the form y = logbx is always increasing if b > 1 and is always decreasing if b < 1.

As with other types of functions, logarithmic functions are translated vertically when a constant is added to the function value and translated horizontally when a constant is added to x within the function. Their scale, or stretch, is affected if the x-value or the function value is multiplied by a constant.

In Lesson 1.4, we
explained how
all functions f(x),
including polynomial
functions, are
translated vertically
k units for f(x) + k
or horizontally h
units for f(xh).

Note that the range
will still be all real
numbers, even if
the graph appears
vertically stretched
or shrunken.

A logarithmic function of the form y = a ⋅ logbc(x − h) + v is the function y = logbx adjusted as follows:

• When |a| > 1, the graph is stretched vertically. When |a| < 1, the graph is shrunk vertically.

• When |c| > 1, the graph is shrunk horizontally. When |c| < 1, the graph is stretched horizontally.

• The graph is shifted horizontally h units: to the right if h is positive (when a constant is subtracted from x), to the left if h is negative (when a constant is added to x).

• The graph is shifted vertically v units: up if v is positive, down if v is negative.

• When a is negative, the graph is reflected across the line y = v (the x-axis when v = 0).

• When c is negative, the graph is reflected across the line x = h (the y-axis when h = 0).

Graph the function p(x) = log4(x − 3) + 5. What are the domain and range of p(x)?

The graph of y = log4x is always increasing and passes through (1, 0) and (4, 1), as shown below.

The graph of p(x) = log4(x − 3) + 5 is the graph of y = log4x shifted 3 units to the right and 5 units up. The vertical asymptote will be moved 3 units to the right, to x = 3. Instead of passing through (1, 0) and (4, 1), p(x) will pass through points 3 units right and 5 units up from each of those: (4, 5) and (7, 6).

Because the graph is shifted 3 units to the right, the domain is now x > 3. This is also apparent from the equation. The argument of the logarithm, (x − 3), must be greater than 0 in order for the logarithm to be defined, so x must be greater than 3. The range, however, is still all real numbers.

No matter how much
you shift the parent
graph y = logbx
vertically or adjust its
scale, its range will still
be all real numbers.

Use technology to graph the function f(x) = 3ex. Algebraically solve for the inverse function, g(x), and graph it.

Here is the graph of f(x) = 3ex.

x = 3ey

Switch x and y in y = 3ex.

x/3 = ey

Divide both sides by 3.

ln(x/3) = y

Take the ln of both sides.

lnx − ln3 = y

Use the logarithmic quotient property to rewrite ln(x/3).

So, g(x) = lnx − ln3. Using technology, we can graph g(x), as shown below. As expected, it is the reflection of f(x) = 3ex across the line y = x.

The value of ln3 is a
constant approximately
equal to 1.1, so the
graph of g(x) is the
graph of lnx (which
is approximately
log2.7x) shifted about
1.1 units down. Even
without graphing
technology, we can
get a sense of what
this graph looks like.

Graph the function s(x) = log3(−x − 2) without technology.

Put this function into the form y = a · logbc(xh) + v. In this case, the value of a is 1 and the value of v is 0, so we just need to write the function in the form y = logbc(xh). The x within the parentheses must be positive, as shown, so c must equal −1.

s(x) = log3−1(x + 2)

Factor out −1 in the argument.

We can graph s(x) based on the graph of y = log3x. The graph of y = log3x is always increasing, passes through (1, 0) and (3, 1), and has a left arm pointing down along a vertical asymptote of the y-axis. It also passes through the point (9, 2), because log39 = 2. Here is the graph of y = log3x.

The graph of s(x) = log3−1(x + 2) is the graph y = log3x translated 2 units to the left and reflected across the vertical line x = −2, as shown below.

In the form
y = logbc(xh), the
value of h in
y = log3−1(x + 2) is
−2. The addition of 2
to x is the same as a
subtraction of −2.