### Video Transcript

Find the solution set of three to the power of π₯ plus 243 over three to the power of π₯ equals 36 in the set of real numbers.

To find the solution set, weβre going to solve this equation for π₯. That is, weβre going to find the values of π₯ that satisfy the equation. But it looks rather not set at the moment, so weβre going to manipulate it a little first. Weβre going to begin by multiplying through by three to the power of π₯. Three to the power of π₯ times three to the power of π₯ is three to the power of π₯ squared. 243 divided by three to the power of π₯ then timesed by three to the power of π₯ is just 243. And 36 times three to the power of π₯ is as shown.

And you might notice this looks a little like a quadratic equation. To make that more clear, weβre going to perform a substitution. Weβll let π¦ be equal to three to the power of π₯. And so, our equation becomes π¦ squared plus 243 equals 36π¦. Now, we know that, to solve a quadratic equation, we need to set it equal to zero. So to achieve this, weβll subtract 36π¦ from both sides. And so, our equation becomes π¦ squared minus 36π¦ plus 243 equals zero.

And we have a number of ways we can solve this quadratic equation. We could complete the square or use the quadratic formula. Alternatively, if the expression on the left-hand side is factorable, we can go down that route. So letβs see. First, we know that the first term in each binomial must be π¦, since π¦ times π¦ gives us π¦ squared. Then, we need to find two numbers whose product is 243 and whose sum is negative 36. Well, these two numbers are negative 27 and negative nine.

Now, we can say that since the product of these two binomials is zero, either one or other of the binomials must themselves be zero. That is, either π¦ minus 27 is zero or π¦ minus nine is zero. Letβs solve this first equation for π¦ by adding 27 to both sides. And when we do, we get π¦ is equal to 27. Similarly, we solve this second equation for π¦ by adding nine to both sides. And we find that π¦ is equal to nine.

The problem is we were looking to solve this in terms of π₯. So, we go back to our earlier substitution π¦ is equal to three to the power of π₯. Replacing π¦ with this value and we see that either three to the power of π₯ is equal to 27 or three to the power of π₯ is equal to nine. We know, though, that three cubed is 27 and three squared is nine. So, π₯ is equal to three or π₯ is equal to two.

And so, we have the solution set of our equation. Itβs three and two.