# Video: Solving Exponential Equations by Factorisation

Find the solution set of 4^(𝑥² + 8) − 1,026 × 2^(𝑥² + 8) + 2,048 = 0 in ℝ.

03:20

### Video Transcript

Find the solution set of four to the power of 𝑥 squared plus eight minus 1,026 times two to the power of 𝑥 squared plus eight plus 2,048 equals zero in the set of real numbers.

Believe it or not, this is actually a special type of quadratic equation. We are going to need to perform some manipulation to make it look like that though. And the key here is spotting that four is the same as two squared. And so, this means we can write four to the power of 𝑥 squared plus eight as two squared to the power of 𝑥 squared plus eight. But since we multiply the exponents in this case, we can reverse this and say that this is the same as two to the power of 𝑥 squared plus eight all squared.

And so, we can rewrite our equation somewhat. We get two to the power of 𝑥 squared plus eight all squared minus 1,026 times two to the power of 𝑥 squared plus eight plus 2,048 equals zero. And then we perform a substitution. We’re going to let 𝑦 be equal to two to the power of 𝑥 squared plus eight. And so our equation simplifies quite a lot. We get 𝑦 squared minus 1,026𝑦 plus 2,048 equals zero. And then we have a number of ways to solve this. Let’s solve by factoring the expression on the left-hand side.

We know that the first term in each binomial must be 𝑦, since 𝑦 times 𝑦 is 𝑦 squared. And then we’re looking for two numbers whose product is 2,048 and whose sum is the coefficient of 𝑦. So that’s negative 1,026. Well, these numbers are negative 1,024 and negative two. And then, since the product of our two binomials is zero, we know that this means at least one of the binomials must themselves be zero. So, either 𝑦 minus 1,024 is zero or 𝑦 minus two is zero.

Let’s solve this first equation for 𝑦 by adding 1,024 to both sides. And we get 𝑦 equals 1,024. Similarly, we’ll add two to both sides of our second equation. So, 𝑦 is equal to two. But remember, we were looking to find the values of 𝑥 which satisfy our original equation. So, we’re going to go back to our substitution where we said 𝑦 is equal to two to the power of 𝑥 squared plus eight. By replacing 𝑦 with two to the power of 𝑥 squared plus eight, we form two new equations in 𝑥. They are two to the power of 𝑥 squared plus eight equals 1,024 and two to the power of 𝑥 squared plus eight equals two.

Well, we actually know that two to the power of 10 is 1,024 and two to the power of one is two. So, this means that in our first equation, 𝑥 squared plus eight must be equal to 10. And in our second, 𝑥 squared plus eight must be equal to two. We’ll solve these equations for 𝑥 by subtracting eight from both sides. So, we get 𝑥 squared equals two in our first equation, and 𝑥 squared equals negative six in our second.

Our second step is to take the positive and negative square root of our equations. In our first equation, we get 𝑥 is equal to the positive and negative square root of two. And in our second, we get 𝑥 is equal to the positive and negative square root of negative six. But we were looking for the solution set in the set of real numbers, and the square root of negative six does not yield a real result. So, there are no solutions that we’re interested in for the equation 𝑥 squared equals negative six. And so, there are two values in our solution set. They are the square root of two and the negative square root of two.