It’s your turn on exponential functions

Hello I’m, Carlo, and together with Alberto, we will guide you along the “Steps in practice”, or “It’s your turn” steps, during the four weeks of this MOOC. If you have already followed the first pre-calculus about numbers and functions, you already know how it works. Well, here we solve some problems whose text is written below in that same web page. It is very important that you watch this video just after having tried to solve the problems by yourself. So in case you didn’t, please stop watching this video and try to solve the problems whose text is written below here. So in that case, today we are dealing with exponentials.

And we are asked to draw the graph of 1/3 to the x from that of 3 to the x. Well, it is important to notice that 1/3 to the x is exactly 1 over 3 to the x, and so this is equal to 3 to minus x. So this means that the value of the function 1/3 to something at x is that of 3 to something at minus x. So we take x and we look at the value of 3 to something at minus x. Well, this is exactly the value at x of 1/3 to something. And so we get this point of the graph– the point x, 1/3 to the x.

And similarly, doing this with the other points, we get the graph of 1/3 to the x. In exercise two, we want to simplify some expressions. Let us begin with 81 to minus 1/4.

Well, let us write that 81 is 9 to the square. So this is 9 to the square to the minus 1/4. And the laws of powers give us that this is 9 to minus 2/4, which is 9 to minus 1/2. At the same time, 9 is 3 to the square, so we write 3 to the square to minus 1/2. And again, here we get 3 to minus 2/2, which is 3 to minus 1, or equivalently, 1/3. The second, 25 to minus 3/2, so 25 to minus 3/2. Well, let us write that 25 is 5 to the square. So this is 5 to the square to minus 3/2.

The laws of powers give us 5 to minus 3, which is 1 over 5 to the 3. And the last one, 16 over 81 to minus 1.25. Well, if you multiply 1.25 by 4, we get 5. So this is 16 over 81 to minus 5/4. Now 16 is 4 to the square, and is also 2 to the 4, and 81 is 9 to the square, but at the same time 3 to the 4th, so here we have 2 to the 4th divided by 3 to the 4th to minus 5/4. So this is actually 2 over 3 the 4 power, and this is at the same time to minus 5/4.

And the laws of powers give us 2 over 3 to minus 5. And equivalently, 2/3 is 3/2 to minus 1, so this is 3/2 to minus 1 to minus 5. And so finally we get to 3/2 to the power of 5, which is also 3 to the 5 divided by 2 to the 5. And this ends exercise two.

In exercise three, we want to find the domain of some exponential functions, or composition with exponential functions. Well, the first one, let us remind that 2 to x is defined whenever x is any real number. So in particular, since 2 x squared plus 5 x is defined for every real number, the same happens to 2 to 2 x squared plus 5 x. So this is defined for x belonging to R. 1/2 to square root of x, well 1/2 to something is defined for every something. And square root of x is defined for x greater or equal than 0. So here, this is defined just for x greater or equal than 0.

The last one, x minus 1 to pi, well let us remind that y to pi is defined for a every y greater or equal than 0. So x minus 1 to the pi is defined whenever x minus 1 is greater or equal than 0. That is x greater or equal than 1.

Now y to the minus 1/2 is defined whenever y is strictly positive. It is not defined for y equal to 0. So 3 minus x to minus 1/2 is defined whenever 3 minus x is strictly positive. That is, x strictly less than 3. Finally 5 to something is defined for every something. But square root of x plus 1 is defined whenever x plus 1 is greater or equal than 0. So 5 to square root of x plus 1 is defined whenever x plus 1 is greater than 0. So whenever x is greater or equal than minus 1. So finally, we have three conditions that we have to put together that they have to be satisfied at the same time.

That is x greater than 1, strictly less than 3, and x greater or equal than minus 1. So finally, we get x greater or equal than 1, and at the same time strictly less than 3. And this is the solution, the domain of the third function, and the solution to exercise three. See you in the next step.