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Video: Solving Equations Involving Positive or Negative Exponents of Variables or Simple Terms

Tim Burnham

A detailed walk-through of a range of examples solving equations with the unknown, or a simple linear expression involving the unknown, to a positive or negative exponent. For example, 𝑥³ = 8, 1/𝑥² = 1/25, (𝑥 − 5)⁴ = 16, or (𝑥 + 3)⁻³ = 27.

16:55

Video Transcript

In this video we’re gonna be solving equations involving positive or negative exponents of variables or simple terms. So we’re gonna be going through a few examples like the one that you can see below here and just going through how to solve them and present your answers.

So let’s start with a nice simple one then. Find the value of 𝑥 when 𝑥 squared is equal to sixteen. So to find the value of 𝑥 when we got the value of 𝑥 squared, we’re gonna have to take square roots because that’s the opposite of squaring.

So again we’re using inverse operations to undo these exponents. So the square root of 𝑥 squared is just 𝑥, but the square root of sixteen? Well four times four is sixteen, but negative four times negative four is also sixteen. So there are two possible values.

And we use this plus or minus symbol to represent the fact that the answer could be positive four or it could be negative four. And when we write out the answer, we’ll actually write them as two separate solutions: the answer is 𝑥 equals four or 𝑥 equals negative four.

Now the thing is questions like this is when the exponent is even, then you’ll always have this positive negative issue. So for even exponents if 𝑥 is negative, you’ll have an even number of negatives cancelling out. So the 𝑥-value could be negative and you get a positive answer or the 𝑥-value could sort of being positive and you’ll still get a positive answer.

Number two: 𝑥 cubed equals eight. Solve for 𝑥. So we’re gonna use the inverse operation again and the-the opposite if you like of cubing something is cube rooting something. So we’re gonna take cube root of both sides. So the cube root of 𝑥 cubed is equal to the cube root of eight. Well the cube root of 𝑥 cubed is just 𝑥 because it’s 𝑥 times 𝑥 times 𝑥 that gives us 𝑥 cubed. And the cube root of eight is two because two times two times two is equal to eight. Now it can’t be negative two because negative two times negative two times negative two — well negative two times negative two will be four and four times negative two will be negative eight. So when you’ve got an odd exponent, then you’re not going to have the positive negative dilemma. So for questions like this, there’s only one answer. And in this case that’s 𝑥 equals two.

Now for number three, we got solve for 𝑥. We got one over 𝑥 squared equals one over twenty-five. Now for this question I’ve got one over 𝑥 squared. So one over something is equal to one over twenty-five is equal to one over something. So if one over something is equal to one over something else, those two things must be equal.

So 𝑥 squared must be equal to twenty-five. So I’m just gonna take square roots of both sides. And the square root of 𝑥 squared is just 𝑥. And the square root of twenty-five remember is an even power, so we are going have to have either the positive or negative version. And five is the square root.

So when we write that out in our answer, 𝑥 is equal to five or 𝑥 is equal to negative five. Or we could even write it out as a solution set. So using set notation, there are two elements in that set: negative five and five.

Now let’s do this question again. But what if I didn’t spot this little trick at the beginning that said that the denominators had to be equal because the numerators were already equal. Well I’ll be looking to get the 𝑥-term off of the denominator. So I’d be multiplying both sides of my equation by 𝑥 squared so that I can cancel out the 𝑥 squared on the left-on the left-hand side.

But because they’re fractions I wouldn’t just multiply it by 𝑥 squared. I’d multiply it by the fraction version of 𝑥 squared, which is just 𝑥 squared over one. Now on the left-hand side I’ve got 𝑥 squared on the numerator and I’ve got 𝑥 squared on the denominator, so they will both cancel out and my left-hand side just becomes one over one, which is one.

And on the right-hand side I’ve got 𝑥 squared on the numerator and twenty-five times one which is just twenty-five on the denominator. Now I need to multiply both sides by twenty-five to try to get rid of that number on the denominator on the right-hand side. And again writing twenty-five in its fraction form on the right-hand side so that I can just cancel those two out and leave myself with 𝑥 squared on the right-hand side.

Well now I’m ready to take square roots of both sides. Well in fact I’m now at the position that I was right at the beginning of the other question, so I can now jump to this point up here. So spotting that little trick at the beginning and equating the denominators saved all of this work over on the right-hand side.

Okay let’s move on to example number four then. Solve for 𝑥. 𝑥 minus five all to the power of four is equal to sixteen. So I’m gonna be taking the fourth root of both sides because taking the fourth root is the inverse operation of taking something to the exponent of four. And the fourth root of 𝑥 minus five all to the power of four is just 𝑥 minus five. And that’s why we did the fourth root of course.

And the fourth root of sixteen, well two times two times two times two is sixteen, so the fourth root is two. But of course remember it’s an even exponent. So minus two or negative two would also work. Negative two times negative two is four, negative two times negative two is four, and four times four is sixteen. So there are two possible answers: positive or negative two.

Well over the left-hand side, I’ve got 𝑥 minus five is equal to positive or negative two. So if I add five to both sides of my equation, I’ll just leave myself with 𝑥 on the left-hand side there. So on the left I got 𝑥 minus five plus five. Well clearly that minus five add five becomes zero. So I’m just left with 𝑥. And on the right-hand side, I’ve either got positive two plus five which is seven or I got negative two plus five which is positive three. So again we got two possible answers.

And depending on where you live and what curriculum you’re following, you’ll probably either present it in this format 𝑥 is equal to seven or 𝑥 is equal to three or you’ll write out your solution set and say that the solution set contains the elements three and seven.

Let’s move on to number five then. So one over 𝑥 plus two all to the power of three is equal to zero point zero zero one and we gotta solve for 𝑥. Well I would strongly recommend that rather than working with decimals on this one, we would work as a fraction.

Now if we think of the place value of these decimal positions: the first decimal is tenth, the second decimal is hundredth, and the third decimal is thousandth. So this means one thousandth, so one over a thousand. So I can simply rewrite that expression for our equation as one over 𝑥 plus two all cubed is equal to one over a thousand.

Now I’m gonna do our old trick again. We’ve got one over something is equal to one over something. So those two things must be equal. And how we got 𝑥 plus two all cubed is equal to a thousand? So I need to think of inverse operations: what’s the inverse operation from cubing something? It’s cube rooting it.

And so the cube root of the left-hand side is equal to the cube root of the right-hand side. Well the cube root of something cubed is just the thing. So on the left-hand side, I’ve just got 𝑥 plus two. That’s what inverse operations do; they cancel each other out. And on the right-hand side, I’m looking for the cube root of a thousand, which is equal to ten. Remember odd powers: you don’t need to worry about the plus or minus thing. So 𝑥 plus two is equal to ten. Now all I need to do is take two away from each side of my equation, so that the two take away two on the left-hand side leaves us with just 𝑥 and ten take away two on the right-hand side leaves us with eight. So we got one answer: 𝑥 equals eight. And our solution set would just look like this and we just have one element in it and that would be eight.

Okay next example, 𝑥 plus three all to the power of negative three is equal to twenty-seven and we’ve gotta solve this for 𝑥. So we’ve got an odd power, so we don’t have to worry about the positive negative thing. But it’s a negative exponent here, negative three, so we need to recall that having a negative exponent means it’s one over the things. That’s gonna be one over 𝑥 plus three all to the power of three. So we’ve just evaluated the negative part of the exponent. We’ve sort of worked out what it means. It means we have to turn that fraction over.

And now I’ve put twenty-seven into its fraction form, twenty-seven over one. Now we’re not quite in the nice situation where we were before, where we had the same numerator of one so we could just compare denominators. So we’re gonna have to multiply both sides of our equation by 𝑥 plus three all to the power of three. We’ve just said that we can remove that variable from the denominator on the left-hand side.

Now you notice that I’ve put the twenty-seven back to adjust the twenty-seven as it doesn’t need to be a fraction. The only reason I tried it as a fraction was to see if we could compare numerators or compare denominators. So on the right-hand side we’ve just got twenty-seven times 𝑥 plus three all cubed. And on the left-hand side what we’ve got- we want the fraction version of 𝑥 plus three all to the power of three so that we can then cancel those out, I’ve got 𝑥 plus three all to the power of three on the bottom. Divide that by itself, you get one, same on the top, you get one. So on the left-hand side now, I’ve just got one times one over one times one and that is just one.

Well now I’ve got twenty-seven times 𝑥 plus three all cubed. So what I’m gonna do is actually divide both sides by twenty-seven so that I can just isolate the 𝑥 plus three all cubed term on its own. A cubed term on its own, the inverse operation is gonna be the cube root. So that’s kind of where we’ve headed with that. So let’s just divide both sides by twenty-seven first.

And that means we can cancel out the twenty-seven’s on the right-hand side. And it means that we’ve got one over twenty-seven is equal to 𝑥 plus three all cubed. So now we’re getting to the point we can do the cube root. So we’re gonna do the cube root of both sides: so the cube root of one over twenty-seven on the left-hand side and the cube root of 𝑥 plus three cubed on the right-hand side. Well obviously that inverse operation is just gonna leave us with 𝑥 plus three on the right-hand side.

But what is the cube root of one over twenty-seven? Well it’s the same as the cube root of one over the cube root of twenty-seven. So I’m just gonna write it that way for now. So I’ve got the cube root of one over the cube root of twenty-seven is equal to 𝑥 plus three. Well the cube root of one is just one because one times one times one is one. And the cube root of twenty-seven is three because three times three times three is nine times three, that’s twenty-seven.

And we’re nearly there. So to isolate 𝑥, I’ve just got to subtract this three from both sides. And that means over on the right-hand side, I’ve got 𝑥 plus three minus three, which is just 𝑥 because those two cancel out. That’s the point of subtracting three from both sides. So now I’ve got to do one-third minus three.

Well step one is when we’re doing fraction subtraction, I need to basically have two fractions. So I’m gonna do one-third minus three over one. And I need to come up with an equivalent fraction for three over one, which has a denominator of three. So I’m gonna multiply it by three over three. Three over three is just one. So we’re not changing the magnitude of the number; we just wanna get an equivalent fraction for three with the same denominator of three.

And three times three on the top is nine; one times three on the bottom is three. So I’ve now got one-third minus nine-thirds. Well this takes us into the awful territory of negative fractions, -and- but one take away nine is negative eight. So I’ve got negative eight-thirds. And if I convert that into a mixed number, I’ve got 𝑥 is equal to negative two and two-thirds.

So it was a bit of a more tricky example there. But because we had an odd power here, we didn’t have to worry about the positive or negative thing. But we did have the negative exponent, which meant that we had to flip the fraction. We then found that we couldn’t take this nice shortcut of comparing denominators because we have one over something is equal to twenty-seven over one. So we had to multiply both sides out as we did here. And then we ran into this additional problem here of the cube root of one over twenty-seven, but we could express that as the cube root of one over the cube root of twenty-seven, which enabled us to work out what that answer was relatively easily.

Well then onto our last example, we’ve got ten thousand over 𝑥 minus four all to the power of negative four equals one and we gotta solve for 𝑥. So we got an even exponent, four, which means we’re gonna have the whole positive and negative two possible answers situation. But we’ve also got a situation where it’s a negative exponent. So we’ve got a negative exponent on the bottom of a fraction. So I’m just going to reinterpret that.

And 𝑥 minus four to the power of negative four on the-on the denominator there is the same as 𝑥 minus four to the power of four on the numerator. That negative exponent means we have to flip that part of the fraction. So that’s moving up to the numerator there. Well I’m trying to get 𝑥 on its own. So I’ve got to see what I can do about this ten thousand term here. Well the inverse operation to multiplying by ten thousand is dividing by ten thousand. So that’s what I’m gonna do to both sides of my equation. So there we have it. We’ve divided both sides by ten thousand. And on the left-hand side, I’ve now got ten thousand on the numerator times by something and ten thousand on the denominator. I can cancel those out. And so I’ve just got 𝑥 minus four all to the power of four on the left- hand side.

And that’s equal to one over ten thousand, so something to the power of four. So the inverse operation to the power of four is taking the fourth root. So we’re gonna take the fourth root of both sides. And the fourth root of 𝑥 minus four all to the power of four is just 𝑥 minus four. Inverse operations, they cancel each other out. And the fourth root of one over ten thousand, I’m gonna rewrite that as the fourth root of one over the fourth root of ten thousand.

And so the fourth root of one is one because one times one times one times one is one. So the numerator is just one and on the denominator the fourth root of ten thousand is just ten because ten times ten is a hundred, times ten is a thousand, times ten is ten thousand. So our equation has become 𝑥 minus four is equal to one over ten.

But whoa whoa whoa! I hear you say hold on! The fourth root of one could be negative one because negative one times negative one times negative one times negative one is one as well. So actually this is plus or minus and this is plus or minus as well. So what we’ve got is 𝑥 minus four is plus or minus a tenth- is positive or negative one-tenth. So now 𝑥 minus four is equal to positive or negative a tenth. And I need to add four to both sides so that I can get 𝑥 on its own on the left-hand side.

So on the left-hand side, the whole point of adding four was that I got negative four plus four which equals zero, so that just leaves me with 𝑥. And on the right-hand side, I’ve got two possible values. So I’ve either got positive a tenth plus four or I’ve got negative a tenth plus four. So that’s either four and a tenth or it’s three and nine-tenths or four point one or three point nine. And then we have out two possible answers.

So what do we have to deal with in that last question then? Well we had the negative exponent which meant that we had to flip the fraction in the first place. And we had the even power, so we had that kind of positive or negative issue which we nearly missed when we were going through the question. So you have to be very very careful of that.

And that meant that in the end we had two possible answers. We also had the issue where we had the fourth root of one over ten thousand, which we split up into the fourth root of one over the fourth root of ten thousand.

So hopefully those examples will help you to solve equations involving positive or negative exponents of variables or simple terms.