### Video Transcript

In this video, weβll learn how to
sketch logarithmic functions with different bases and their transformations, and
weβll study their different characteristics. So, letβs begin by reminding
ourselves what we actually mean when we call a function logarithmic.

Now, a logarithmic function is the
inverse of an exponential function. Itβs of the form π of π₯ is equal
to log base π of π₯, where π is greater than zero and not equal to one. And we say if the point π₯, π¦
satisfies the exponential function, then the point π¦, π₯ satisfies the logarithmic
function. And the fact that the exponential
and logarithmic functions are inverses of one another is actually really helpful
when it comes to sketching their graphs. And we will investigate that in a
little more detail later on in this video. For now, letβs begin by
investigating the shape of the curve given by a logarithmic function.

Find the missing table values for β
of π₯ equals log base two of π₯.

And then we have a table with three
missing values. So, what do we mean when we talk
about a logarithm? Logarithmic functions are inverses
to exponential functions. Imagine we have the given
expression. We read this as log base π of π
equals π. π is the base, π is set to be the
exponent in this expression, and π is called the argument. And in fact, this is exactly the
same way as describing the relationship between π, π, and π as π to the power of
π equals π. So with this in mind, letβs take
the function log base two of π₯, and weβll take our first value of π₯, negative
two.

Substituting π₯ equals negative two
in the function for β of π₯ gives us β of negative two is equal to log base two of
negative two. But we need to work out the value
of β of negative two. So letβs call that π sub one. If we equivalently describe this
relationship as two to the power of π sub one equals negative two, we see we need
to find the value of π sub one that satisfies this equation. But there is no power of two that
will give us a value of negative two. This could in fact only be achieved
if the base itself was negative. So, π sub one is in fact not
defined. β of negative two then, we say, is
undefined.

Letβs now move on to π₯ equals
one. β of one in our table will be the
value of log base two of one. Defining β of one to be equal to π
sub two this time, we can equivalently write this relation as two to the power of π
sub two equals one. To solve this equation, we ask
ourselves, what power of two gives an answer of one? Well, in fact, the only way for
this to be true is if π sub two equals zero. Any real number not equal to the
power of zero raised to the power of zero will always be one. So β of one and the second value in
our table is zero.

Weβre now going to repeat this for
π₯ equals two. β of two is log base two of
two. Defining β of two as π sub three,
we see that we can represent this relation as two to the power of π sub three
equals two. So once again, we ask ourselves,
what power of two gives two? Well, the only power of two that
will give us two itself will be one. So π sub three, and in fact β of
two, is equal to one. And so weβve completed the values
in our table; they are undefined, zero, and one, respectively.

We might now look to sketch the
graph of this function. And to do so, weβll need to work
out a few more values. And we could, of course, use the
same method as before. Alternatively, we could simply type
these into a calculator. Letβs suppose π₯ is equal to
four. Log base two of four is equal to
two. Similarly, if π₯ is equal to eight,
we get log base two of eight, which is equal to three.

Now, β of negative two is
undefined. In fact, the function itself is
undefined for any values of π₯ less than or equal to zero. So, we have an asymptote here, an
asymptote of the line π₯ equals zero or the π¦-axis. So, the graph of our function β of
π₯ equals log base two of π₯ looks a little bit as shown. The domain, we said, is the values
of π₯ we can substitute in, and thatβs purely positive real numbers: π₯ is greater
than zero or π₯ is contained in the open interval from zero to β. We also observe that the graph
itself passes through the π₯-axis at π₯ equals one. And we can generalize these two
properties for graphs of logarithmic functions.

Consider graphs of the form π¦
equals log base π of π₯, where π is greater than zero and not equal to one. They have just one
π₯-intercept. They pass through the π₯-axis at
one. In fact, they will also pass
through the point π, one. In our last example, the graph of
π¦ equals log base two of π₯ passes through the point two, one. These graphs all have a vertical
asymptote given by the π¦-axis or π₯ equals zero. Finally, these functions have a
domain where π₯ is in the open interval zero to β and a range where the output, the
π¦-values, are in the open interval from negative β to β. So it looks a little something like
this.

Now, in fact, just as with
exponential graphs, the size of π can tell us a little bit more about the
function. If π is greater than one, then the
function itself is increasing. And if π is greater than zero or
less than one, the function is decreasing. Now in fact, we might even look to
compare the graph of this function with its inverse, π¦ is equal to π to the power
of π₯. The graph of π¦ equals π to the
power of π₯ looks a little something like this. As we might expect when we draw the
graph of a function and its inverse, they will be reflections of each other in the
line π¦ equals π₯. Now this can be a really useful
fact if weβre struggling to remember what either graph looks like. So now letβs have a look at a
question that involves identifying the graph of a logarithmic function.

Which graph represents the function
π of π₯ equals log base five of π₯?

And we have five graphs to choose
from. So, letβs begin by inspecting the
function weβve been given. Itβs a logarithmic function, and it
has the general form of the logarithmic function π of π₯ equals log base π of π₯
where, of course, our value of π, which is here five, cannot be equal to one and is
greater than zero. Now, one of the features we know
about this function is that it passes through the π₯-axis at one, but it also passes
through the point π, one. So weβre looking for a graph which
passes through one, zero and five, one. We also know that the π¦-axis or
the line π₯ equals zero is an asymptote to this graph. In other words, the graph of our
function approaches the π¦-axis but never actually reaches it. And we know that when π is greater
than one, the graph itself is purely increasing. Itβs increasing over the entire
domain of the function.

Well, in fact, all of our graphs,
if we look carefully, are increasing and they all have the π¦-axis as an
asymptote. So we need to identify which of our
graphs pass through the point one, zero and five, one. In fact, if we plot each of these
points on every single graph, we see that only one graph passes through both
points. The correct graph, then, is
(A). Graph (A) represents the function
π of π₯ equals log base five of π₯.

Letβs now identify the graph of a
second logarithmic function.

Which curve is π¦ equals log base
three of π₯?

And then we have a coordinate plane
with four individual graphs drawn on it. Well, here we have a logarithmic
function, a function of the form log base π of π₯, where π is positive and not
equal to one. One of the properties of this
function is that its graph passes through the point one, zero and the point π,
one. So our graph, since π is equal to
three, will pass through one, zero and three, one. We also know that if our value of
π is greater than one, then the graph is increasing over its entire domain.

And this is really useful because
we can actually disregard options (c) and (d) straightaway. We notice that those graphs, whilst
they do pass through the point one, zero, are decreasing over their entire
domain. And so they could be a
transformation via reflection of a logarithmic graph or a logarithmic graph with a
fractional base. But they certainly arenβt the curve
of π¦ equals log base three of π₯. Finally, we know that all of these
curves should have an asymptote given by the line π₯ equals zero or the π¦-axis. Well, all of our graphs do have
this asymptote. We observe that the graphs seem to
approach the π¦-axis but never quite reach it.

So, we need to find the curve then
that passes through the point one, zero and three, one. We have already shown that all
curves pass through the point one, zero, but the only curve that also passes through
the point three, one is curve (a). So, curve (a) is the graph of the
equation π¦ equals log base three of π₯.

So at this point, weβve considered
how to generate a group of ordered pairs using a table of function values, and weβve
identified a couple of logarithmic graphs. It might be clear, though, that we
wonβt always be working with logarithmic functions on their own. Sometimes we might be dealing with
composite functions or a function of a function. In these scenarios, it can be
helpful to recall what we know about function transformations to help us identify
the relevant graph. Letβs demonstrate one such
transformation in our next example.

Which function represents the
following graph? Is it (A) π of π₯ equals log base
four of two π₯? (B) π of π₯ equals log base four
of π₯. Is it (C) π of π₯ equals log base
two of two π₯? Is it (D) π of π₯ equals log base
two of π₯? Or is it (e) π of π₯ equals log
base eight of π₯?

Weβll begin by inspecting the graph
of the function itself. It certainly resembles the graph of
a logarithmic function. We have a vertical asymptote of the
π¦-axis. The graph certainly seems to
approach this line but never quite reaches it, and the function itself is
increasing. The domain is values of π₯ greater
than zero and the range appears to be all real numbers. And so itβs likely that our
function is of the form π of π₯ equals log base π of π₯. And, in fact, all of the equations
that weβve been given are of this general form, although some of them have two π₯ as
their argument.

So one of the things we know about
the function π of π₯ equals log base π of π₯ for positive values of π not equal
to one is that they pass through the point one, zero and the point π, one. By plotting the point one, zero on
the graph of our function, we see that it doesnβt actually pass through this
point. In fact, it passes through the
point one-half, zero, so it looks like it could be a transformation of our original
function. We might then recall that say we
have a function π¦ equals π of π₯, the function π¦ equals π of ππ₯, where π is
not to be confused with the base in log base π of π₯, gives a horizontal
compression but by a scale factor of one over π.

So, if we had the function π of π₯
equals log base π of π₯, the function π of π₯ equals log base π of two π₯ would
be a compression by a scale factor of one-half in that horizontal direction. So, this means that our graph could
be (A) or (C). It could be log base four of two π₯
or log base two of two π₯. Letβs choose a point on the curve
to deduce whether itβs π of π₯ equals log base four of two π₯ or π of π₯ equals
log base two of two π₯. We see our curve passes through the
point eight, two and the point two, one. For due diligence, weβll check both
of these coordinates in our individual functions.

When π₯ is equal to two, the first
function, (A), gives us π of two equals log base four of two times two. Thatβs log base four of four, which
we see is equal to one. So, this point satisfies the
function π of π₯ equals log base four of two π₯. Then, π of eight is log base four
of two times eight. Thatβs log base four of 16. And we know four squared is equal
to 16, so this must be equal to two. Once again, our second point eight,
two satisfies this function, so we can deduce that the function is π of π₯ equals
log base four of two π₯.

Letβs check, though, by
substituting two and eight into our second function. When π₯ is equal to two, the
function is log base two of four, which is two. And when π₯ is equal to eight, the
function is log base two of 16, which is equal to four. Neither point two, two nor eight,
four lie on the graph weβve been given, and so we can confirm that it cannot be
option (C); itβs (A).

Weβve now demonstrated how to
identify and sketch graphs of logarithmic functions as well as looked at their
various transformations. So, letβs finish up by recapping
some of the key points from the lesson. In this lesson, we saw that the
logarithmic function is the inverse of an exponential function. Itβs of the form π of π₯ equals
log base π of π₯, where π is positive and not equal to one. Their graphs pass through the point
one, zero and π, one. They have an asymptote given by the
π¦-axis or the line π₯ equals zero, and the value of π can tell us about its
shape. If π is greater than one, the
graph is increasing over its entire domain. And if π is between zero and one,
it is decreasing.