Video: Solving Exponential Equations by Factorization

Find the solution set of 2^(2π‘₯ βˆ’ 2) βˆ’ 130 Γ— 2^(π‘₯ βˆ’ 1) + 256 = 0.

03:40

Video Transcript

Find the solution set of two to the power of two π‘₯ minus two minus 130 times two to the power of π‘₯ minus one plus 256 equals zero.

Now, it may not look like it, but what we have here is a special type of quadratic equation. We’re going to need to find a way to manipulate it a little bit. We’re going to need to recall one of our laws of exponents. And that is, when we divide two numbers whose base is the same, we subtract their exponents. The converse is, of course, also true. So, π‘₯ to the power of π‘Ž minus 𝑏 equals π‘₯ to the power of π‘Ž divided by π‘₯ to the power of 𝑏. And so, what this means we can write two to the power of two π‘₯ minus two as two to the power of two π‘₯ divided by two to the power of two. We write two to the power of π‘₯ minus one as two to the power of π‘₯ divided by two to the power of one. But of course, two to the power of one is just two. So, this is two to the power of π‘₯ divided by two. And so, we have our equation.

Now, the next thing we need to do is try to simplify our equation by getting rid of the fractions. To achieve this, we’re going to multiply through by the highest power of two, by two squared. Well, two to the power of two π‘₯ divided by two squared times two squared is two to the power of two π‘₯. Then, our second term becomes negative 130 times two to the power of π‘₯ times two. Then, 256 times two squared is 1024. Now, of course, multiplication is commutative. So, we can rewrite our second term as negative 130 times two times two to the power of π‘₯, which is negative 260 times two to the power of π‘₯.

And then, we recall another one of our laws of exponents. And that is, π‘₯ to the power of π‘Ž all to the power of 𝑏 is the same as π‘₯ to the power of π‘Ž times 𝑏. And so, the converse must be true. And we can rewrite two to the power of two π‘₯ as two to the power of π‘₯ all squared. And we’re now ready to make our substitution. We’re going to let 𝑦 be equal to two to the power of π‘₯. Then, two to the power of π‘₯ all squared is 𝑦 squared. Negative 260 times two to the power of π‘₯ is negative 260𝑦. And our equation is, therefore, 𝑦 squared minus 260𝑦 plus 1024 equals zero.

So, how do we solve this equation? Well, one method we have is to factor the expression 𝑦 squared minus 260𝑦 plus 1024. To do so, we write it as a product of two binomials. The first term in each binomial is 𝑦, since 𝑦 times 𝑦 is 𝑦 squared. And then, we need to find two terms whose product is 1024 and whose sum is negative 260. Well, these values are negative four and negative 256. And so, our equation becomes 𝑦 minus four times 𝑦 minus 256 equals zero.

But we know that 𝑦 minus four and 𝑦 minus 256 are simply numbers. And these numbers have a product of zero. So, for this to be true, either one of those numbers must themselves be equal to zero. So either 𝑦 minus four is equal to zero or 𝑦 minus 256 is equal to zero. Let’s solve this first equation for 𝑦 by adding four to both sides. So, 𝑦 is equal to four. Similarly, we solve the second equation by adding 256 to both sides. And we get 𝑦 equals 256.

And that’s all fine and well. But we were trying to solve an equation in π‘₯. And so, we go back to our earlier substitution. That was 𝑦 is equal to two to the power of π‘₯. By replacing 𝑦 with this expression, we get two equations in π‘₯. It’s two to the power of π‘₯ equals four and two to the power of π‘₯ equals 256. Now, we could use logs to solve each of these equations. But actually, these are exponents of two that we should know by heart. The exponent of two that gives us four is two and the exponent of two that gives us 256 is eight.

And so, now, we’ve solved for π‘₯. We can use these little curly brackets to represent the set of solutions to our equation. And when we do, we find that the solution set for our equation is two and eight.

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