### Video Transcript

Find the solution set of five to the π₯ power plus 125 over five to the π₯ power equals 30.

What I want to do first is treat five π₯ like itβs its own variable. And that means I wanna get five π₯ out of the denominator. To do that, weβll multiply the entire equation by five π₯. Five π₯ times five π₯ equals five to the two π₯. Five π₯ times 125 over five π₯ equals 125. 30 times five π₯ equals 30 times five π₯.

What we want to do now is set this equation equal to zero. Weβre going to subtract 30 times five π₯ from both sides of the equation. Our equation now says five to the two π₯ minus 30 times five to the π₯ plus 125 equals zero.

Weβre going to take this equation and factor it. We want to say five π₯ plus something times five π₯ plus something else. In this case, our constant value is positive, and our term in the middle is negative. That tells me that when Iβm factoring, Iβm going to be dealing with subtraction on both terms.

I need two values that multiply together to equal 125 and that add together to equal 30. 25 times five equals 125, and 25 plus five equals 30. So we can set each of these values equal to zero: five π₯ minus five equals zero and five π₯ minus 25 equals zero. On the left, add five to both sides. We need the π₯ value such that five π₯ equals five. π₯ equals one. Five to the first power equals five.

Our next equation, we add 25 to both sides. We need an π₯ value such that five to the π₯ power equals 25. Here π₯ has to be equal to two. Our solution set π₯ could be one or two, our solution set one, two.

You might be looking at this factorization and thinking, βWhat on Earth did you just do there?β I want to show you something that will hopefully clear this up. Imagine our equation said π plus one over 25π equals 30.

The first thing I would want to do is multiply the entire equation by π. π times π equals π squared. π times 125 over π equals 125, and π times 30 equals 30π. To find π, I wanna set the entire equation equal to zero. So we subtract 30π from both sides. Now we have π squared minus 30π plus 125 equals zero. We can factor this π squared. If we factor it, we find that π minus five times π minus 25 is equal to this statement.

We say π minus five equals zero and π minus 25 equals zero. π equals five and π equals 25. But our π value, our variable, was five to the π₯ power. We needed five to the π₯ power equals five, and five to the π₯ power equals 25. And thatβs what I meant at the beginning when I said we want to consider five to the π₯ power as its own variable, as itβs a variable in itself. When you do that, you end up with coefficients that are very easy to factor and get the solution set of one and two.