### Video Transcript

Use the following graph to find the
solution set of the equation two to the power π₯ minus three equals π₯ minus
two.

When we are given a question like
this involving an equation and several functions, there will be a relationship
between the equations and the functions. Itβs very common that the left-hand
side of the equation is linked to one function. Letβs call that function π sub one
of π₯. The right-hand side of the equation
is usually another function. Letβs call that π sub two of
π₯. The expressions on both sides of
the equation will either be the exact functions or something very closely
related. The left-hand side of this equation
has an exponent. So if there was the function π sub
one of π₯ equals two to the power of π₯ minus three, it would produce an exponential
function.

The blue line on this graph has the
shape of an exponential function. So we will need to check if this
function could be written as π sub one of π₯ equals two to the power π₯ minus
three. We will also need to check if this
green straight line can be written as the function π sub two of π₯ equals π₯ minus
two. We can investigate this green
straight line first. Recall that we can graph the
function π of π₯ equals π₯ in this way. Itβs a straight line with a
gradient of one and a π¦-intercept of zero.

Using this line to draw the
function π of π₯ equals π₯ minus two, we perform a vertical translation of π of π₯
equals π₯ by two units downwards. The function π of π₯ equals π₯
minus two is parallel, so it has the same gradient of one. But the π¦-intercept will be
negative two. And so we have established that the
green line is indeed a function, which we could denote as π sub two of π₯ equals π₯
minus two.

We can check the second function by
inputting some coordinates on the curve. The simplest way to do this is by
finding some coordinates with integer values. For example, we could observe that
the coordinate five, four lies on the curve. If we substitute π₯ equals five
into the function, letβs see if we get an output or π of π₯ value of four. So when π₯ is equal to five, we
have π sub one of π₯ equals two to the power five minus three. Thatβs two squared. And of course, two squared is equal
to four. This means that the function takes
an input of five and produces an output of four.

We can then check another
coordinate. The coordinate four, two also lies
on the curve. So when we substitute π₯ equals
four into the function, we get π sub one of π₯ equals two to the power four minus
three. Simplifying this gives two to the
power one, which of course is simply two. So once again, an input π₯ into the
function produces an output of two. And so we can be satisfied that
this function can be represented by π sub one of π₯ equals two to the power π₯
minus three. This function is also a horizontal
translation of the function π of π₯ equals two to the power π₯.

Now we know that we have these two
functions, we can solve the given equation. To find the solution set of this
equation, we need to begin by finding any common points on both functions. These points are the points of
intersection. We can see that there are two
points of intersection, the coordinates three, one and four, two. The solution set will be the
π₯-values of each of these coordinates. Therefore, we can give the answer
that the solution to the equation two to the power π₯ minus three equals π₯ minus
two is the set containing three and four.