Video: Solving a System of Exponential Equations

Find the solution set of 216^(𝑥 + 𝑦) = 1296 and 3^(𝑥 − 3𝑦 − 1) = 81.

05:37

Video Transcript

Find the solution set of 216 to the power of 𝑥 plus 𝑦 equals 1,296 and three to the power of 𝑥 minus three 𝑦 minus one equals 81.

So what we’re gonna do here is form a pair of simultaneous equations to help us find 𝑥 and 𝑦. I’m going to start with the one that involves 216 to the power of 𝑥 plus 𝑦 equals 1,296. So what we want to do in this kind of question is we want to have the same base number on the left- and right-hand side of our equation. So we take a look at 216 and 1,296. And we could see that 216 can be written in an exponent form because it can be written as six to the power of three. And 1,296 can be written as six to the power of four.

So we can rewrite our equation as six to the power of three to the power of 𝑥 plus 𝑦 equals six to the power of four. So it’s at this point we can use one of our exponent rules. So we’ve got 𝑥 to the power of 𝑎 to the power of 𝑏, which is equal to 𝑥 to the power of 𝑎𝑏. So we multiplied the exponents. So what we’re gonna get is six to the power of three 𝑥 plus three 𝑦 equals six to the power of four. And it’s worth noting that we’ve got a left-hand side because what I did is I multiplied three by 𝑥 plus 𝑦. So to do this, what I did was I distributed across the parentheses and got three 𝑥 plus three 𝑦.

Now, if we look at this, we can see that we’ve got the same base on the left- and right-hand sides. So what we can do is equate the exponents. So I’ve now got three 𝑥 plus three 𝑦 equals four. And this is the first of our simultaneous equations. So I have labeled it number one. Okay, great. Let’s take a look at the other equation to see if we can form another simultaneous equation. So now what I’ve got is three to the power of 𝑥 minus three 𝑦 minus one equals 81. And we can see here already that we’ve got the base number because we’ve got three as the base on the left-hand side. And 81 can be written with the base three because we can say it’s three to the power of four.

So now what we have is three to the power of 𝑥 minus three 𝑦 minus one equals three to the power of four. And once again, we’ve got the same base on the left- and right-hand side of the equation. So what we can do is we can equate the exponents. And when we do that, we’ve got 𝑥 minus three 𝑦 minus one equals four. And if we add one to each side of the equation, we get 𝑥 minus three 𝑦 equals five 𝑦 [five]. So we can call this our second equation or equation two. Okay, great. So we’ve now got our two equations. Let’s solve them simultaneously to find 𝑥 and 𝑦.

Now, with our pair of simultaneous equations, we can use one of two methods. We could either use substitution or we could use elimination to solve. Well, we’re gonna use the elimination method. And we’ll do that because we can see that, in both of our equations, we’ve got three 𝑦. Okay, one is positive and one is negative, but because we’ve got the same number as our coefficient, what we can do is we can use elimination. And elimination is gonna work because what we can do is we can add equations one and equation two. And this is gonna eliminate our 𝑦s.

So what we’re gonna get is four 𝑥 and that’s cause three 𝑥 plus 𝑥 is four 𝑥. And then, we’ve got positive three 𝑦 add negative three 𝑦, which just gives us zero, and then four add five, which gives us nine. So we’ve now got an equation four 𝑥 equals nine. So we’ve removed one of our variables 𝑦. And now we’ve just got an equation of 𝑥, so we can solve this to find 𝑥.

So what we can do is divide both sides of the equation by four, and we get 𝑥 is equal to nine over four or nine-quarters. Okay, great. So now, let’s find out what 𝑦 is. So now what we’re gonna do is we’re gonna substitute 𝑥 equals nine over four into equation two. You can substitute it into either equation to find 𝑦; it’ll work either way. I’m just gonna substitute it into equation two. And when we do that, we get nine over four minus three 𝑦 equals five. And then next, what we’re gonna do is we subtract five and add three 𝑦 to each side of the equation. And when we do that, we get nine over four minus five equals three 𝑦.

Well, we can rewrite five as 20 over four. And that’s because if we think about it, there’ll gonna be twenty-quarters in five holes. So we’ve now got nine over four minus 20 over four equals three 𝑦, which is gonna give us 11 over four equals three 𝑦. So one more stage to find 𝑦 and that is to divide by three. So if we divide both sides of our equation by three, we get 11 over 12 equals 𝑦. It’s worth noting here that the way we got 11 over 12 is because if we’re dividing 11 over four by three is the same as dividing 11 over four by three over one.

Well, if we’re dividing a fraction, we find the reciprocal of the second faction and multiply. So it’s the same as 11 over four multiplied by one over three. So then we multiply the numerators and the dominators, and we get 11 over 12. It’s key to point out here though that even though we found 𝑥 and 𝑦, we haven’t finished the question because it says “find the solution set,” so we need to leave it in the relevant notation.

So the way we’re gonna leave it is using our set notations. We’ve got our curly brackets, which shows it’s a set. And then, I’ve left it in coordinate form. So we’ve got nine over four and 11 over 12. Nine over four represents our 𝑥-value and 11 over 12 represents our 𝑦-value.

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