### Video Transcript

Find the solution set of 36 to the power of π₯ plus one minus six to the power of π₯ plus five minus six to the power of π₯ plus 216 equals zero.

Weβre looking to solve by finding the values of π₯ which satisfy our equation. And whilst this equation might look really nasty, there are a few ways that we can manipulate it. Firstly, weβre going to use the fact that π₯ to the π times π₯ to the power of π is equal to π₯ to the power of π plus π. In other words, when multiplying two numbers whose base is equal, we add their exponents. And this means we can write 36 to the power of π₯ plus one as 36 to the power of π₯ times 36 to the power of one or just 36 to the power of π₯ times 36.

Similarly, we can write six to the power of π₯ plus five as six to the power of π₯ times six to the power of five or six to the power of π₯ times 7776. And so our equation becomes 36 times 36 to the power of π₯ minus 7776 times six to the power of π₯ minus six to the power of π₯ plus 216 equals zero. But of course, 36 is six squared, so we can write 36 to the power of π₯ as six squared to the power of π₯. But in this case, we can multiply the exponents. So we can further rewrite this as six to the power of two π₯, which we could then write as six to the power of π₯ squared.

Next, we factor negative six to the power of π₯. And our second term is negative six to the power of π₯ times 7776 plus one. Of course, 7776 plus one is 7777. And so our equation is as shown. Now this looks a little bit like a quadratic equation. To make this more clear, letβs perform a substitution. Weβre going to let π¦ be equal to six to the power of π₯. And so our equation becomes 36π¦ squared minus 7777π¦ plus 216 equals zero. And we have a number of ways that we can solve this equation. One method is to factor the expression on the left-hand side. We could also use completing the square or the quadratic formula.

Weβre actually going to factor this. And this might take a little bit of trial and error. But when we do factor, we get 36π¦ minus one times π¦ minus 216 equals zero. Now, for this to be true, for the product of these two binomials to be equal to zero, either one or other of the binomials must themselves be zero. So 36π¦ minus one is either equal to zero or π¦ minus 216 is equal to zero. Letβs solve this first equation for π¦ by adding one to both sides, to get 36π¦ equals one. Then we divide through by 36. And we see that a solution to this equation is π¦ equals one over 36.

The second equation is a little easier to solve. We add 216 to both sides such that π¦ is equal to 216. But of course, we said that to find the solution set, we need to find the values of π₯ which satisfy our equation. So weβre going to go back to our earlier substitution π¦ equals six to the power of π₯. And we can say that six to the power of π₯ is equal to one over 36 or six to the power of π₯ equals 216. Now, for this first equation, the only value of π₯ which makes this true is negative two. And for our second equation, itβs three. And so we have two solutions in our solution set. They are negative two and three.