### Video Transcript

Rearrange the equation π¦ equals two to the power of π₯ plus one minus one over two to the power of π₯ plus one to find π₯ in terms of π¦. Hence, determine the inverse of π to the function π of π₯ equals two to the power of π₯ plus one minus one over two to the power of π₯ plus one.

The first part of this question asks us to make π₯ the subject. And so the first thing weβre going to do is multiply both sides by the denominator of our fraction, by two to the power of π₯ plus one. On the left-hand side, we get π¦ to the power of two π₯ plus one. And on the right, weβre just left with the numerator of our fraction. Thatβs two to the power of π₯ plus one minus one.

Next, weβre going to distribute our parentheses by multiplying two to the power of π₯ by π¦ and one by π¦. That gives us two to the power of π₯ times π¦ plus π¦ equals two to the power of π₯ plus one minus one. Now remember, weβre looking to make π₯ the subject. So, weβre going to begin by getting all the π₯βs onto one side of our equation.

Letβs subtract two to the power of π₯ plus one from both sides so that two to the power of π₯ times π¦ plus π¦ minus two to the power of π₯ plus one equals negative one. Next, weβll subtract π¦ from both sides. And so we see that two to the power of π₯ times π¦ minus two to the power of π₯ plus one equals negative one minus π¦.

And here we recall one of our laws of exponents. This says that π₯ to the power of π times π₯ to the power of π is the same as π₯ to the power of π plus π. And this means we can write two to the power of π₯ plus one as two to the power of π₯ times two to the power of one, or just two to the power of π₯ times two.

Next, we factor that expression on the left-hand side. So we get two to the power of π₯ times π¦ minus two. And our next step is to divide through by π¦ minus two. And we get two to the power of π₯ equals negative one minus π¦ over π¦ minus two.

Now itβs important to realize that had we at this step instead subtracted two to the power of π₯ times π¦ from both sides, we would actually have had one plus π¦ over two minus π¦. These expressions are the same. Theyβre equivalent fractions. And theyβre retrieved by multiplying the numerator and denominator of one of our fractions by negative one.

Our final step is to take the log base two of both sides of our equation. Now the reason this step is useful is because one of the laws of logs tells us that the log base two of two to the power of π₯ is the same as π₯ times the log base two of two. But log base two of two is simply one. So weβve successfully written π₯ in terms of π¦. π₯ is log base two of negative one minus π¦ over π¦ minus two. So how does this help us find the inverse of the function π?

Well, one way we can find the inverse is by swapping π₯ and π¦. So we would write this as π₯ equals two to the power of π¦ plus one minus one over two to the power of π¦ plus one. We then rearrange this formula to make π¦ the subject. Now, of course, we did this a slightly different way. We simply rearranged to make π₯ the subject. So we know that how we rearrange this second equation, we would have got π¦ is equal to log base two of negative one minus π₯ over π₯ minus two. And that means this is the inverse of our function π. The inverse π of π₯ is equal to log base two of negative one minus π₯ over π₯ minus two.