 In this video I want to talk about solving equations involving power functions and specifically we're going to look at power functions So the form x to the m over in power So we're only going to use We're only going to use rational exponents in the situation and also we're going to assume that these rational exponents are written in Lowest terms so m and n have no common divisors amongst them now One thing to remember here is that when you take x to the m over n This is the same thing by exponent rules as the nth root of x to the m So radicals and powers might be involved in this situation So imagine first we're in the situation where m over n right here where our denominator Our denominator here is an odd power So if your denominator is odd that means you're talking about things like the third root or the fifth root or the seventh root These functions have no restrictions on their domains whatsoever We we only have to worry about the even roots and so the domain in this situation is going to be all real numbers and Likewise The range is a little bit more complicated because that's a lot to do with the numerator here in which case you see something like the following if the numerator If the numerator turns out to be an odd number as well So like an example would be like three-fifths if you had something like that then we're going to see that our range Is going to be all real numbers and our function is going to be always increasing It's gonna be always increasing on the other hand if your numerator turns out to be an even number because that is allowed in that situation your range would actually only be Zero to infinity because like we saw with even monomials before if you square a real number It cannot be negative and so if your numerator is is even that means you've essentially squared your function And therefore you can't get negatives and it won't always be increasing. It'll actually be decreasing and on the interval negative infinity to zero and then it'll be increasing on the interval Zero to infinity So in that regard this blue part The red part it kind of looks like an even monomial and the blue part kind of looks like an odd monomial So if your denominator is odd the numerator being even or odd It'll kind of it will look a lot like an a monomial in some regard And that's not exactly true because the concavity can be very very different, right? So an odd monomial will look something like this, but we could get things like the following The concavity could be totally whack. We're not gonna worry about that right now Now in the situation where x equals the exponent m over n Let's suppose that the denominator is actually an even number in that situation Now of course if the denominator is even and since the fractions in lowest terms that means the numerator must necessarily be odd So there's only one situation to consider here But there's there is an important caveat in this situation your domain is not gonna be all real numbers Your domain is only gonna be zero to infinity Right because if you take a negative at for x since the since you're taking a square root because if you have an even denominator You're taking a square root or a fourth root or a sixth root You have to worry about imaginary numbers and if x is negative You're gonna take the square root of an image of an a negative number, which is imaginary So its domain we're gonna restrict it to just be zero to infinity Otherwise we'd have to allow for Imaginary solutions which that's possible in terms of graphing we're not gonna allow that but for solving algebraic equations We're gonna allow that well. What does it do to the range? Well, it turns out in that situation that whatever the situations you're in before right when you are going from zero to infinity This function will be increasing and so we see the same thing there your range will look like Zero to infinity and you'll see that this thing is always increasing now. I could look something like this But it could also look something like this there's a couple possibilities Now what is why does the graph of things matter so much as you're trying to solve these things? So it does play an important role right here. Let me switch the color Let's look at the equation x to the five force is equal to 32 Now one the way to solve equations like this is going to use the following property If you have x to the m and then you raise that to the n that actually means you can multiply the powers together That was one of our exponent rules So when you want to solve an equation like x to the five force here What you're going to do is you're going to raise both sides of the equation to the reciprocal exponent So notice that we have the five fourths power We're going to raise that to the four fifths power To try to solve this thing because then on the left hand side You're going to get x to the first which is just x Five fourths times four fifths, but then this equals 32 to the four fifths Which you want to think that as the fifth root of 32 to the fourth In which case you get that the fifth root of 32 is two since two 32 is just two to the fifth Then raise that to the fourth you're going to get 16 So it looks like the solution to this equation here is going to be 16 And that is perfectly correct. There's no issue with that And the issue that you have to kind of watch out for is the following first of all Notice we're raising both sides to the four fifths power because the denominator is odd This is going to be a one to one function. It has a true inverse things are going to be hunky and dory But when we switch to say the next example, we have to watch out for things Notice we have x the two fifths power is equal to 64 To solve this one we take the two fifths power and we're going to be taking the reciprocal power of both sides Which the reciprocal power here is going to be five halves take five halves Which on the left hand side the two fifths power and the five half power cancel each other out And on the right hand side notice what we have here We're going to have the square root of 64 raised to the fifth power Now because we're taking the square root that actually means there's two solutions here We have to have a plus or minus And this is why I was talking about the issues about domain and range earlier about increasing decreasing Because this has to do with whether the function is actually invertible or not If your numerator here is an even number Then you're going to need a plus or minus with your solution Because in order to get rid of a square you have to take a square root But there's actually two possibilities because an even monomial is a two to one function And so we see here that the solutions would actually be plus or minus Uh the square root of eight, sorry square root of 64 is eight So we have to take eight to the fifth which will be 32,768 But in particular there's two solutions here. There's plus 32,768 and minus 32,608 These are things we have to watch out for but another issue that we have to watch out for is the following What if we had something like Like this so we had x to the two fifths is equal to negative 64 In this situation when we take the square root of both sides we'll end up with x is equal to Plus or minus the square root of negative 64 to the fifth Which that actually gives us an imaginary number, right? So we get plus or minus 332,768 I as your two solutions and so You have to whether you have a positive or negative makes a big difference here If we want if we want real solutions We'd say there's no solution right here for allowing complex solutions Then we'd say something like this So when your numerator is even some funky things happen when the numerator is even Then the then you'll get two solutions for which They could be imaginary solutions. So you have to watch out for so treat even numerators with caution We didn't have that issue when we did odd numerators if the denominator was even there's no big deal whatsoever with this following exception if you had If you took the situation where x to the five fourths was equal to negative 32 In this situation you're going to get no solution going on here Because like we talked about earlier the range of this function is zero to infinity. So if your denominator is Even then that cannot equal a negative number even though you might think there's solution There isn't one that's outside the range of the function. So if your numerator Is even you'll have two solutions and they could be non real solutions If your denominator is even then the right hand side have better be positive. Otherwise, there's no solution Um, now the thing is if both numerator and denominator are odd You don't have to worry about these issues. So in some respect even numbers are the odd behaving ones I know it's a horrible pun, but it's the truth Um, let's take something like this. Let's take the equation three equals x to the three fourths is equal to x to one half Now in this situation, we have x is on both sides of the equations. How can we deal with that? How can we get rid of how can we get rid of the the rational exponents here? Well, if you were trying to add something like three fourths plus a half What you would probably do is like, well, I guess I need to find a common denominator Three fourths plus two fourths, and then you add those together to get five fourths. We want to kind of do the same thing here We're going to take our function three x to the three fourths And we're going to take a power on both sides, but we want to take the least common denominator So that's going to be four in this situation. So I'm going to take both sides of the equation and raise it to the fourth power I guess what I'm trying to say is if you had something like the following if you had three fourths x plus Um Seven is equal to like one half x. How could you deal with this equation? You could simplify things by kind of times in both sides of the equation by four This would clean up the fractions entirely. You get three x plus 28 is equal to two x We want to kind of clear the same we want to clear denominators for the exponents We have to look for least common denominators here. So we take the fourth power on both sides By exponent rules the left hand side would become three to the fourth I'll switch that back to white three to the fourth times x cubed and the the right hand side would look like x squared That's what we have here and then we would try to simplify this thing We get 81 x cubed is equal to x squared. How do you solve a polynomial equation like this? It turns out that the way we solved it for quadratics works the same way here We could try to factor it 81 x cube minus x squared equals zero You can factor out an x squared on the left hand side giving you 81 x minus one And then by the zero product property, we must have that either x squared equals zero or 81 x minus one equals zero Solving the first one tells you x equals plus or minus zero, which is actually the same number zero Solving the other one says you get one over 81 and those would be the two solutions to this equation right here So when you have rational exponents, sometimes we try to clear the denominators In this case, we use exponential properties as opposed to multiplying both sides by the same number