### Video Transcript

Solving Exponential Equations
Graphically

In this video, we will learn how to
determine the number of solutions to an exponential equation given graphically, and
we will also see how to apply this to solve exponential equations using graphical
methods. Before we begin by trying to solve
exponential equations graphically, letβs start by recalling what we mean by an
exponential function.

We recall that an exponential
function is one of the form π of π₯ is equal to π multiplied by π to the power of
π₯, where π is positive and π is not allowed to be equal to one. Well, itβs worth noting these are
not the only example of exponential functions. For example, we could multiply our
value of π₯ by a constant, or we could add a constant onto this value of π₯, and
this would still be an exponential function. However, for now, weβll just focus
on this type of exponential function. This then means that an exponential
equation is an equation involving an exponential function. So, thatβs one of the form π times
π to the power of π₯ is equal to some function π of π₯. A solution to this equation would
be a value of π₯ which makes both sides of the equation equal.

Sometimes we can solve equations by
using algebraic manipulation. However, for exponential equations,
this is often very difficult. This is because π₯, our variable,
appears in the exponent. So instead, weβll focus on finding
these solutions graphically. To see how we might solve an
exponential equation graphically, letβs start with an example. Letβs assume we were given the
graph of the exponential function π¦ is equal to four to the power of π₯, and letβs
say we were asked to solve the equation four to the power of π₯ is equal to one. This means we need to find the
value of π₯ we substitute into our exponential function to output a value of
one. We can do this by using our
graph. Remember, the π¦-coordinate of
every point on our curve tells us the output value for our function at that value of
π₯.

We want to know when our function
outputs a value of one, it will output a value of one when its π¦-coordinate is
equal to one. So, we sketch the line π¦ is equal
to one onto our axes. We can then see when our curve has
a π¦-value of one. It has a π¦-value of one when its
π₯-value is equal to zero. In other words, weβve shown thereβs
one point of intersection between the line π¦ is equal to one and the curve π¦ is
equal to four to the power of π₯. Thatβs the point with coordinates
zero, one. At this value of π₯, both our
function four to the power of π₯ and the function one output the same value. They output the value of one. Therefore, this must be a solution
to our equation. π₯ equals zero solves the equation
four to the power of π₯ is equal to one.

There are a few other things we can
notice. For example, this is the only point
of intersection between our line and curve. And every solution to our equation
will be a point of intersection between the line and curve. So, because thereβs only one point
of intersection between the line and the curve, we can conclude thereβs only one
solution to this equation. Itβs also worth noting we can check
that π₯ is equal to zero is a solution to our equation by substituting π₯ is equal
to zero into both sides of the equation and making sure theyβre equal. Letβs start with substituting π₯ is
equal to zero into the left-hand side of the equation.

Substituting π₯ is equal to zero
into the left-hand side of our equation, we get four to the power of zero. And we can evaluate this by using
our laws of exponents. We know any nonzero number raised
to the zeroth power is always equal to one. We can then do the same with the
right-hand side of our equation. However, the right-hand side of
this equation is just a constant value of one, so the value of π₯ does not affect
this value. Therefore, when π₯ is equal to
zero, both the left- and right-hand side of our equation are equal. This confirms that π₯ is equal to
zero is a solution to our equation.

We can use this exact same method
to solve other exponential equations. For example, letβs solve the
equation four to the power of π₯ is equal to five minus π₯. Once again, since a solution to our
equation is a value of π₯ such that both sides of our equation are equal, we can
find the solutions to our equation by looking for points of intersection between the
curve π¦ is equal to four to the power of π₯ and the line π¦ is equal to five minus
π₯ because the points of intersection will have the same output for both of our
functions, in other words, they will be solutions to our equation.

We already have a graph of π¦ is
equal to four to the power of π₯. So, on the same axis, we should
sketch the line π¦ is equal to five minus π₯. First, we know that its
π¦-intercept is five. We can also find the π₯-intercept
of this line. We substitute π¦ is equal to zero
and solve for our value of π₯. We see that the π₯-intercept of
this line is when π₯ is equal to five. We can use this to plot the
line. We know its π¦-intercept is at five
and its π₯-intercept is also at five. Then the straight line connecting
these two points is the line π¦ is equal to five minus π₯.

Finally, we can see that thereβs
one point of intersection between our curve and our line, and this will be the point
where the outputs of both of these functions are equal. We can see that the π₯-coordinate
of this point of intersection is one. So, we have that π₯ is equal to one
is a solution to our equation. And in fact, since this is the only
point of intersection between the line and the curve, this is the only solution to
our equation. However, we do need to be
careful. We sketch the line π¦ is equal to
five minus π₯, and we use this to estimate the point of intersection between the
line and the curve. So, we canβt be sure that π₯ is
equal to one is the exact solution to our equation since we are approximating by
using our sketch.

To show that π₯ is equal to one is
the solution to this equation, weβre going to need to substitute π₯ is equal to one
into both the left- and right-hand side of our equation and check if theyβre
equal. We can start with the left-hand
side of our equation. Substituting π₯ is equal to one, we
get four to the first power. And by using our laws of exponents,
we know any number raised to the first power is equal to itself. So, four to the first power is
equal to four. We can then do the same with the
right-hand side of our equation. Substituting π₯ is equal to one, we
get five minus one, which we can evaluate is equal to four. Therefore, since both the left- and
right-hand side of our equation are equal when π₯ is equal to one, we can conclude
that π₯ is equal to one is a solution to our exponential equation.

So far, all of our equations have
had solutions. However, it is also possible that
an equation has no solutions. For example, we can see that the
line π¦ is equal to negative two and the curve π¦ is equal to four to the power of
π₯ have no points of intersection. This means if we were asked to
solve the equation four to the power of π₯ is equal to negative two by using the
given diagram, we would be able to conclude that there are no solutions to this
equation because any solution to this equation would be a point of intersection
between the curve π¦ is equal to four to the power of π₯ and the line π¦ is equal to
negative two. And instead of saying there are no
solutions to this equation, we can introduce the idea of the solution set.

The solution set of an equation is
the set of all solutions to that equation. So, instead of saying the equation
four to the power of π₯ is equal to negative two has no solutions, we can say that
its solution set is the empty set. Letβs now go through an example
where weβre given the graph of an exponential function and we need to use this to
determine the solution set of an exponential equation.

Use the given graph of the function
π of π₯ is equal to two to the power of five minus π₯ to find the solution set of
the equation two to the power of five minus π₯ is equal to two.

In this question, weβre given the
graph of an exponential function and this exponential function appears in the given
exponential equation. We need to use this to determine
the solution set of the equation. First, we recall the solution set
of an equation is the set of all solutions to that equation. Therefore, weβre looking for the
set of all values of π₯ which balance both sides of the equation. Another way of thinking about this
is since two to the power of five minus π₯ is equal to the function π of π₯, we can
substitute π of π₯ into our equation. This gives us the equation π of π₯
is equal to two. Weβre looking for the set of all
values of π₯ such that π of π₯ is equal to two.

To find these values of π₯, we can
recall that every single point on the curve π¦ is equal to π of π₯ will have
coordinates of the form π₯, π of π₯. In other words, the π¦-coordinates
of the points on the curve tell us the outputs of our function for the given value
of π₯. We want to determine the values of
π₯ where our function outputs two. These will be the points on our
curve with π¦-coordinate equal to two. So, we can find these by sketching
the line π¦ is equal to two onto the same set of axes. We can see thereβs only one point
on our curve of π¦-coordinate equal to two. It will be the point of
intersection between the line π¦ is equal to two and the curve π¦ is equal to two to
the power of five minus π₯. The π¦-coordinate of this point is
two and its π₯-coordinate is four. In other words, when π₯ is equal to
four, our function outputs two. π evaluated at four is two.

Therefore, π₯ is equal to four is a
solution to our equation. In fact, since this is the only
point of intersection between the line and the curve, this is the only solution to
our equation. This means the solution set to our
equation is just the set containing four.

Itβs also worth noting we can check
our answer by substituting π₯ is equal to four into our equation or into our
function. Substituting π₯ is equal to four
into our function π of π₯, we get π evaluated at four is two to the power of five
minus four. Five minus four is equal to
one. So, this simplifies to give us two
to the first power. And any number raised to the first
power is just equal to itself. So, π evaluated at four is equal
to two, which is exactly the same as the right-hand side of our equation, confirming
that π₯ is equal to four is a solution to our equation. Therefore, we were able to show the
solution set of the equation two to the power of five minus π₯ is equal to two is
just the set containing four.

Letβs now go for another example
where weβre given the graph of an exponential function, and we need to use this to
solve an exponential equation.

The diagram shows the graph of π
of π₯ is equal to two to the power of two π₯. Use this graph to find the solution
set of the equation two to the power of two π₯ is equal to four.

In this question, weβre given the
graph of an exponential function, and weβre asked to solve an exponential equation
where this function appears. To do this, we start by recalling
the solution set of an equation is the set of all solutions to that equation. This means weβre looking for all of
the values of π₯ which solve the equation two to the power of two π₯ is equal to
four. To help us do this, letβs start by
replacing two to the power of two π₯ in our equation with π of π₯. This means the equation weβre asked
to solve can be rewritten as π of π₯ is equal to four.

In other words, weβre just looking
for the values of π₯ such that our function π outputs a value of four. And we can recall that the
π¦-coordinate of a point on our curve tells us the output value of a function for
that value of π₯. So, we can find all of the values
where our function outputs four by sketching the line π¦ is equal to four onto our
diagram. And we can then see on our diagram
thereβs only one point on our curve with π¦-coordinate equal to four. This is the point of coordinates
one, four.

And itβs worth reiterating this
tells us that π evaluated at one is equal to four. And therefore, one is a solution to
our equation. In fact, all solutions to our
equation will be a point of intersection between the line π¦ is equal to four and
the curve π¦ is equal to two to the power of two π₯. So, because we can see thereβs only
one point of intersection, we know thereβs only one solution. Therefore, the solution set of the
equation two to the power of two π₯ is just the set containing one.

Letβs now see an example where we
first need to rearrange the exponential equation that weβre given.

The diagram shows the graph of π
of π₯ is equal to two to the power of π₯ over two. Use this graph to find the solution
set of the equation two to the power of π₯ over two plus five is equal to nine.

In this question, weβre given the
graph of an exponential function π of π₯. And weβre asked to use this to
determine the solution set of an equation which contains our function π of π₯. To do this, we start by recalling
the solution set of an equation is the set of all solutions to that equation. In this case, it will be the set of
all values of π₯, such that two to the power of π₯ over two plus five is equal to
nine. To answer this question, it can
help us to rewrite our exponential equation in terms of the function π of π₯. Substituting two to the power of π₯
over two is equal to π of π₯ into our equation, we get π of π₯ plus five is equal
to nine. We can simplify this equation
further by subtracting five from both sides. We get π of π₯ is equal to nine
minus five, which simplifies to give us π of π₯ is equal to four. So, we want to find the values of
π₯ such that our function outputs a value of four.

Remember that the π¦-coordinate of
any point on our curve tells us the output value of our function at that value of
π₯. So, we want to find all of the
points on our curve with π¦-coordinate four. We do this by sketching the line π¦
is equal to four onto our diagram. We can see thereβs only one point
of intersection between our line and our curve. And we can see that this point has
π₯-coordinate four. Therefore, when we input a value of
π₯ is equal to four into our function, the output value is four. π of four is equal to four. And in fact, since this is the only
point of intersection between our line and our curve, this is the only solution to
our equation. Therefore, the solution set of this
equation is the set containing four.

We can check that π₯ is equal to
four is a solution to our equation by substituting π₯ is equal to four into the
left-hand side of our equation. Substituting π₯ is equal to four
into the left-hand side of our equation, we get two to the power four over two plus
five, which we can simplify four over two is equal to two. So, this is equal to two squared
plus five. And then we can evaluate this. Two squared is equal to four. So, we get four plus five, which is
equal to nine, which we can see is exactly equal to the right-hand side of this
equation. Therefore, four is a solution to
our equation, and we know itβs the only solution. Therefore, the solution set of the
equation two to the power of π₯ over two plus five is equal to nine is the set
containing four.

Letβs now see an example where our
exponential equation involves a linear function.

Use the graphs below to answer the
following question. True or false: the equation two to
the power of π₯ is equal to negative π₯ has no solution.

In this question, weβre given the
graph of two functions. Letβs start by determining which
two functions these are the graphs of. First, we can see that our straight
line passes through the origin, so its π¦-intercept is zero. Next, we can see for every one unit
we go across, we travel one unit down. So, its slope is negative one. In slopeβintercept form, thatβs the
line π¦ is equal to negative one π₯ plus zero, which is just π¦ is equal to negative
π₯. Our other curve has the shape of an
exponential function, and we can see it passes through the point with coordinates
one, two. If we substitute π₯ is equal to one
into the function two to the power of π₯, we can see this outputs a value of
two. We could do this with other points
on our curve to conclude that this is indeed a sketch of the curve π¦ is equal to
two to the power of π₯.

We need to use these graphs to
determine whether or not the equation two to the power of π₯ is equal to negative π₯
has a solution. We might be tempted to try and
solve this by using manipulation. However, this will be very
difficult because π₯ appears in the exponent and not in the exponent. Instead, recall that a solution to
this equation is a value of π₯ such that both sides of the equation are equal. In other words, we need to input a
value of π₯ into the function two to the power of π₯ and then put the same value
into the function negative π₯ to get the same output. We can do this directly from our
graph. For the outputs of these two
functions to be equal with the same π₯-input, they must have a point of
intersection. This is because the π¦-coordinate
tells us the outputs of this function for the given input.

Hence, because thereβs one point of
intersection between the line and the curve, we can conclude that two to the power
of π₯ is equal to negative π₯ has one solution. In fact, we can even approximate
this value by trying to read off its π₯-coordinate from the graph. Doing this, we would get that π₯ is
approximately equal to negative 0.6. Therefore, we were able to show
that it is false that the equation two to the power of π₯ is equal to negative π₯
has no solutions.

In our final example, weβll solve
an exponential equation graphically by also sketching a linear function on the same
given graph.

The following graph shows the
function π sub one of π₯ is equal to two to the power of negative π₯. Use this graph and plot the
function π sub two of π₯ is equal to π₯ plus three to find the solution set of the
equation two to the power of negative π₯ is equal to π₯ plus three.

In this question, weβre given two
functions π sub one of π₯ and π sub two of π₯, and weβre given a graph of the
function π¦ is equal to π sub one of π₯. Weβre asked to find the solution
set of an equation. And since π sub one of π₯ is equal
to the left-hand side of this equation and π sub two of π₯ is equal to the
right-hand side of this equation, the equation is π sub one of π₯ equals π sub two
of π₯. We can solve this equation
graphically. Any solution to this equation will
be a point of intersection between the curve π¦ is equal to π sub one of π₯ and the
line π¦ is equal to π sub two of to π₯. Because the point of intersection
would have the same π¦-coordinate and the π¦- coordinate is the output of the
function for the given π₯ coordinator, which means the outputs of the function would
be the same, so our equation would be solved.

We need to sketch the curve π¦ is
equal to π₯ plus three. First, we note that its
π¦-intercept will be at three. We can also find its π₯-intercept
by substituting π¦ is equal to zero. Solving this, we get that π₯ is
equal to negative three. We can then plot our line. Its π¦-intercept is at three, and
its π₯-intercept is at negative three. This then allows us to plot our
line. We just connect the π¦- and
π₯-intercept with a straight line. Then, the only point of
intersection between our line and our curve will be the only solution to our
equation. We can read off its π₯-coordinate;
its π₯-coordinate is negative one.

Then, since the question ask us to
write this as a solution set, weβll write this as the set containing negative
one. Therefore, we were able to show the
solution set of the equation two to the power of negative π₯ is equal to π₯ plus
three is just the set containing negative one.

Letβs now go over some of the key
points of this video. First, the solution set of an
equation is the set of all solutions to that equation, in other words, is the set of
all values which satisfy the equation. And in particular, if an equation
has no solutions, we can say the solution set is the empty set. Next, we saw that we can solve the
equation π of π₯ is equal to π of π₯ by finding the π₯-coordinates of all of the
points of intersection between the graphs π¦ is equal to π of π₯ and π¦ is equal to
π of π₯.

Every point of intersection is a
solution to the equation. And if there are no points of
intersection, then there are no solutions to the equation. Finally, we saw the graphical
solutions to equations can be approximations. This is particularly true if we
need to sketch one of the functions ourselves. In these cases, we should use the
grid lines to try and make our approximation as accurate as possible.