Question Video: Determining the Solution of an Exponential Function Mathematics

Use the graph to find the solution set of the equation 2^(π‘₯ βˆ’ 3) = π‘₯ βˆ’ 2.

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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. It’s very common that the left-hand side of the equation is linked to one function. Let’s call that function 𝑓 sub one of π‘₯. The right-hand side of the equation is usually another function. Let’s call that 𝑓 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 𝑓 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 𝑓 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 𝑓 sub two of π‘₯ equals π‘₯ minus two. We can investigate this green straight line first. Recall that we can graph the function 𝑓 of π‘₯ equals π‘₯ in this way. It’s a straight line with a gradient of one and a 𝑦-intercept of zero.

Using this line to draw the function 𝑓 of π‘₯ equals π‘₯ minus two, we perform a vertical translation of 𝑓 of π‘₯ equals π‘₯ by two units downwards. The function 𝑓 of π‘₯ equals π‘₯ minus two is parallel, so it has the same gradient of one. But the 𝑦-intercept will be negative two. And so we have established that the green line is indeed a function, which we could denote as 𝑓 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, let’s see if we get an output or 𝑓 of π‘₯ value of four. So when π‘₯ is equal to five, we have 𝑓 sub one of π‘₯ equals two to the power five minus three. That’s 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 𝑓 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 𝑓 sub one of π‘₯ equals two to the power π‘₯ minus three. This function is also a horizontal translation of the function 𝑓 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.

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