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Question Video: Solving Exponential Equations by Factorization

Find the solution set of 2^(π‘₯) + 2^(9 βˆ’ π‘₯) = 72 in ℝ.

03:26

Video Transcript

Find the solution set of two to the power of π‘₯ plus two to the power of nine minus π‘₯ equals 72 in the set of real numbers.

Now, at first glance, this equation might look really difficult to solve. So, what we’re going to do is perform some manipulation and that will help clarify what we need to do. Firstly, we know that when we divide two numbers whose base is equal, we can subtract their exponents. And so, the converse must be true. π‘₯ to the power of π‘Ž minus 𝑏 must be equal to π‘₯ to the power of π‘Ž divided by π‘₯ to the power of 𝑏. And what this means is we can rewrite two to the power of nine minus π‘₯ as two to the power of nine divided by two to the power of π‘₯.

Our next job is to simplify our equation by getting rid of the fraction. To achieve this, we multiply through by two to the power of π‘₯. Two to the power of π‘₯ times two to the power of π‘₯ or two to the power of π‘₯ times itself is two to the power of π‘₯ all squared. And we’re going to leave it like that for a moment. Two to the power of nine divided by two to the power of π‘₯ and then timesed by two to the power of π‘₯ is just two to the power of nine, which is 512. And then, 72 times two to the power of π‘₯ is 72 times two to the power of π‘₯. So, we have two π‘₯ all squared plus 512 equals 72 times two to the power of π‘₯.

Now, believe it or not, what we’re actually dealing with is a special type of quadratic. It may not look like it, but we will manipulate it a little further so it does. We begin by subtracting 72 times two to the power of π‘₯ from both sides of our equation. So, we get two to the power of π‘₯ all squared minus 72 times two to the power of π‘₯ plus 512 equals zero. And you might be thinking, β€œThis still doesn’t look like a quadratic equation.” But what we can do is perform a substitution. We’re going to let 𝑦 be equal to two to the power of π‘₯. Then, we can rewrite two to the power of π‘₯ squared as 𝑦 squared and negative 72 times two to the power of π‘₯ as negative 72𝑦.

And so, our equation becomes 𝑦 squared minus 72𝑦 plus 512 equals zero. So, how do we solve this equation for 𝑦? Well, we’re going to factor our quadratic expression. We can write it as the product of two binomials, whose first term is 𝑦. And that’s because 𝑦 times 𝑦 gives us 𝑦 squared. And then, we need to find two numbers whose product is 512 and whose sum is negative 72. Well, these two numbers are negative 64 and negative eight. So, we have 𝑦 minus 64 times 𝑦 minus eight equals zero. And if we think about it, 𝑦 minus 64 and 𝑦 minus eight are just numbers. And these numbers have a product of zero. So, what does that tell us about our individual numbers?

Well, it tells us that either one of these numbers, at least one of them, must be equal to zero. So, either 𝑦 minus 64 equals zero or 𝑦 minus eight equals zero. We can then solve these equations for 𝑦. By adding 64 to both sides of our first, we get 𝑦 equals 64. And by adding eight to both sides of our second, we get 𝑦 equals eight. We really aren’t finished, though. Remember, we were solving an equation in π‘₯, not an equation in 𝑦. And so, we go back to our substitution.

We said 𝑦 is equal to two to the power of π‘₯. So, we can form two new equations and say that either two to the power of π‘₯ equals 64 and two to the power of π‘₯ equals eight. And we might use logs to solve these. But actually, these are powers of two that we should know by heart. We know that two to the power of six is 64 and two to the power of three is eight, or two cubed is eight. So, π‘₯ must be equal to six or three. We use these curly brackets to represent the solution set.

And we find the solution set of two to the power of π‘₯ plus two to the power of nine minus π‘₯ equals 72 is six and three.

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