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Use the following graph to find the solution set of the equation two to the power π₯ minus three equals π₯ minus two.
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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.
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Itβs very common that the left-hand side of the equation is linked to one function.
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Letβs call that function π sub one of π₯.
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The right-hand side of the equation is usually another function.
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Letβs call that π sub two of π₯.
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The expressions on both sides of the equation will either be the exact functions or something very closely related.
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The left-hand side of this equation has an exponent.
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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.
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The blue line on this graph has the shape of an exponential function.
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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.
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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.
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We can investigate this green straight line first.
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Recall that we can graph the function π of π₯ equals π₯ in this way.
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Itβs a straight line with a gradient of one and a π¦-intercept of zero.
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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.
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The function π of π₯ equals π₯ minus two is parallel, so it has the same gradient of one.
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But the π¦-intercept will be negative two.
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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.
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We can check the second function by inputting some coordinates on the curve.
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The simplest way to do this is by finding some coordinates with integer values.
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For example, we could observe that the coordinate five, four lies on the curve.
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If we substitute π₯ equals five into the function, letβs see if we get an output or π of π₯ value of four.
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So when π₯ is equal to five, we have π sub one of π₯ equals two to the power five minus three.
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Thatβs two squared.
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And of course, two squared is equal to four.
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This means that the function takes an input of five and produces an output of four.
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We can then check another coordinate.
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The coordinate four, two also lies on the curve.
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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.
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Simplifying this gives two to the power one, which of course is simply two.
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So once again, an input π₯ into the function produces an output of two.
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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.
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This function is also a horizontal translation of the function π of π₯ equals two to the power π₯.
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Now we know that we have these two functions, we can solve the given equation.
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To find the solution set of this equation, we need to begin by finding any common points on both functions.
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These points are the points of intersection.
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We can see that there are two points of intersection, the coordinates three, one and four, two.
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The solution set will be the π₯-values of each of these coordinates.
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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.