### Video Transcript

Find the solution set of four to the power of π₯ plus two to the power of π₯ plus eight equals 516 in the set of real numbers.

Now, this might look like a really complicated equation to solve for π₯. But the key to solving it is in spotting that four is equal to two squared. This means four to the power of π₯ is the same as two squared to the power of π₯. And we know that, with the two and the π₯, we can multiply these. So we get two to the power of two π₯. Then what weβre going to do is write this as two π₯ all squared. Similarly, weβre going to use our laws of exponents to write two to the power of π₯ plus eight as two to the power of π₯ times two to the power of eight. And these are the laws of exponents weβve used to rewrite our terms.

Now, multiplication is commutative. It can be done in any order. So weβre going to write two to the power of π₯ times two to the power of eight as two to the power of eight times two to the power of π₯. And then we replace four to the power of π₯ with two π₯ all squared. So our equation becomes two to the power of π₯ all squared plus two to the power of eight times two to the power of π₯ equals 516. And then we might spot we have something that looks a little like a quadratic equation. So weβre going to perform a substitution. Weβre going to let π¦ be equal to two to the power of π₯. Then two to the power of π₯ all squared is π¦ squared. Two to the power of eight times two to the power of π₯ is two to the power of eight times π¦. And this is equal to 516.

Well, we can now replace two to the power of eight with 256π¦. And we see that, to solve a quadratic equation of this form, we need to subtract 516 from both sides to make it equal to zero. So we get π¦ squared plus 256π¦ minus 516 equals zero. Now that we have a quadratic equation which is equal to zero, we know that we can solve this by factoring the quadratic expression. We can write it as the product of two binomials. The first term in each binomial must be π¦ because we got π¦ times π¦, which is π¦ squared. And then weβre looking for two numbers whose product is negative 516 and whose sum is 256, the coefficient of π¦.

Well, the only way to achieve this is with negative two and 258. Their product is negative 516, and their sum is positive 256. And so we find that π¦ minus two times π¦ plus 258 is equal to zero. Well, the only way for this to be true is if either π¦ minus two is equal to zero or π¦ plus 258 is equal to zero. We solve this first equation for π¦ by adding two to both sides and this second by subtracting 258. So π¦ is either equal to two or negative 258. But letβs go back to our earlier substitution. We said that π¦ is equal to two to the power of π₯. And so we replace π¦ with two to the power of π₯. And we now have two equations in π₯ that we can solve.

Now, usually when solving equations of this form, we might look to take logs of both sides. But the only power of two that will give us an answer of two is one. And thereβs no value of π₯ such that two to the power of π₯ is equal to negative some constant. And so this second equation has no solutions in the set of real numbers. We can use these curly brackets to represent the solution set. And thereβs only one number in this set; itβs one.