# Video: Finding the Solution Set of Exponential Equations over the Set of Real Numbers Using Factorization and Laws of Exponents

Find the solution set of 2^(𝑥²) = 4^(6𝑥 − 16) in ℝ.

04:10

### Video Transcript

Find the solution set of two to the power of 𝑥 squared equals four to the power of six 𝑥 minus 16 in the set of real numbers.

In order to solve this problem, first of all, we want to have a quick look at the right-hand side of the equation. We can see that the right-hand side of the question is four to the power of six 𝑥 minus 16. And a key bit of information to help us with this problem is that four is equal to two to the power of two or two squared. And the reason this is interesting is because we want to actually get both sides of our equations involving the same base number. And in this case, it’s two. So it’s great since as we know that four is equal to two to the power of two, what we can do is we can actually substitute this into our equation.

So therefore, our equation now becomes: two to the power of 𝑥 squared is equal to two to the power of — and then we have two, because it’s two to the power of two that gives us our four, and then that’s multiplied by six 𝑥 minus 16. So great, we’ve actually achieved what we were setting out to do. And that’s actually to get the same base on both sides of our equations. And as we have the same base on each side of our equation, what we can actually do now is we can actually equate the exponents.

So we get that 𝑥 squared is equal to two multiplied by six 𝑥 minus 16. Great, so we’ve now got an equation. And let’s solve to find 𝑥. So the first thing we’ll do to solve the equation is actually expand the parentheses. So I start with two multiplied by six 𝑥 which gives me 12𝑥. And then I’m gonna have two multiplied by negative 16 which gives me negative 32. So great, my equation will now be 𝑥 squared is equal to 12𝑥 minus 32. As we’re actually looking to solve this equation and find 𝑥, what I’m now gonna do is I’m actually gonna subtract 12𝑥 and add 32 to each side. So what we actually have, our equation equal to zero.

So now we have a quadratic equation which is equal to zero. We have 𝑥 squared minus 12𝑥 plus 32 is equal to zero. And on inspection of this quadratic, we can see that we can factor it to solve it. So we get 𝑥 minus four multiplied by 𝑥 minus eight is equal to zero. So then we got this because the sum of negative four and negative eight is negative 12 which is our coefficient of 𝑥. And the product of negative four and negative eight is equal to positive 32. So, great. Okay, they’re our factors. And now, let’s find 𝑥.

So now as we want to find 𝑥, in order to do that, what we need to do is set both our parentheses equal to zero. Because in order for the equation to equal zero, then one of our parentheses will also have to equal zero. So first, well we’re gonna start with 𝑥 minus four is equal to zero. So add four to each side of the equation. We’re gonna get 𝑥 is equal to four. So great, that’s our first solution.

Then we move across to 𝑥 minus eight is equal to zero. So we just add eight to each side of the equation. And there we have 𝑥 is equal to eight. Great, so that’s the other solution of our equation.

So therefore, we can say that the solution set of two to the power of 𝑥 squared is equal to four to the power of six 𝑥 minus 16 in the set of real numbers is: eight, four.

And just a quick reminder, the key part about a question like this is to make sure that we get both sides of the equation in the same base. And in order to do that, look out for a question where you have, say, a two and a four or a two and an eight, a two and a 16 or a three and nine, a three and a 27. So keep an eye, as that’s the kind of thing that you’ll see in a question like this. And that’s how you’d start the question and move on to solve and find the solution set.