Lesson Video: Solving Exponential Equations Using Exponent Properties Mathematics

In this video, we will learn how to solve exponential equations using the properties of exponents.

17:04

Video Transcript

In this lesson, we’ll learn how to solve exponential equations using the properties of exponents. We are particularly interested in the rules for working with exponents when finding the product, quotient, or power in addition to how we evaluate zero, negative, and fractional exponents. So let’s recall these first. The first rule is the rule for multiplying numbers with exponents. Note that this only works if the base, that’s the big number which here is the π‘₯, is the same. If this is the case, when we multiply these terms, we add their exponents. So, π‘₯ to the power of π‘Ž times π‘₯ to the power of 𝑏 is π‘₯ to the power of π‘Ž plus 𝑏.

Similarly, if we divide these types of numbers, we subtract their exponents. π‘₯ to the power of π‘Ž divided by π‘₯ to the power of 𝑏 is π‘₯ to the power of π‘Ž minus 𝑏. When we’re working with parentheses, in other words, we’re raising an exponential term to another exponent, we multiply these. So, π‘₯ to the power of π‘Ž to the power of 𝑏 is the same as π‘₯ to the power of π‘Ž times 𝑏. We might also recall that anything to the power of zero is one. A negative exponent tells us to find the reciprocal, so π‘₯ to the power of negative π‘Ž is one over π‘₯ to the power of π‘Ž. And a fractional power tells us to find a root, so π‘₯ to the power of one over π‘Ž is the same as the π‘Žth root of π‘₯.

Now, in this lesson, we’re going to use these rules to solve equations involving exponents. And the best way to see how this might look is to consider an example.

Given that two to the π‘₯th power is equal to 32, find the value of π‘₯.

Here we have an equation involving an exponent which is a variable. The exponent here is equal to π‘₯. Now, usually, to solve an equation, we’d look to perform a series of inverse operations, but the inverse to finding the π‘₯th power is to find the π‘₯th root, which doesn’t really help us much. Instead, it’s worth noticing that 32 can be written as a power of two. In fact, we know that two to the fifth power is equal to 32. And this means we can rewrite our equation as two to the π‘₯th power equals two to the fifth power.

So, how does this help? Well, now the base β€” that’s the big number; here that’s two β€” is the same. And so we can say that for this equation to make sense, the exponents must also be the same. That is, π‘₯ is equal to five. Now, whenever we’re solving equations, it always makes sense to check our answer by substituting it back into the original equation. Let’s let π‘₯ be equal to five. And then two to the π‘₯th power becomes two to the fifth power, which is indeed equal to 32. And so, given that two to the π‘₯th power is equal to 32, π‘₯ must be equal to five.

So, that’s a fairly straightforward example of how to solve exponential equations. What we’re going to do next is incorporate some of the rules of exponents in an example.

Given that eight to the power of 𝑦 equals four to the power of 𝑧 which equals 64, find the value of 𝑦 plus 𝑧.

So this is a rather unusual-looking equation because it has three parts to it. Now, the key to answering this problem is to spot that each of the numerical parts of our equation can be written as a power of some common number. In fact, they can be written as two to the power of something. The first few powers of two are shown. We know two squared is four, two cubed is eight, all the way up to two to the sixth power is equal to 64. So, let’s replace each numerical part of our equation with its associated power of two. Eight is two cubed. So, eight to the power of 𝑦 is two cubed to the power of 𝑦. Then, four is two squared, so four to the power of 𝑧 is two squared to the power of 𝑧. And finally, 64 is two to the sixth power.

But how does this help us? Well, if we recall a rule for working with powers of exponents, we know that π‘₯ to the power of π‘Ž to the power of 𝑏 is equal to π‘₯ the power of π‘Ž times 𝑏. Here we multiply the exponents. And so, two cubed to the power of 𝑦 can be written as two to the power of three times 𝑦 or just two to the power of three 𝑦. And then two squared to the power of 𝑧 is the same as two to the power of two 𝑧. And so, our equation is now two to the power of three 𝑦 equals two to the power of two 𝑧 which equals two to the power of six.

Now that the base, that’s the big number which is here two, is the same, we can say that for this equation to make sense, the exponents themselves must also be the same. In other words, three 𝑦 must be equal to two 𝑧, which must in turn be equal to six. Now, we can actually split this equation up to find individually the value of 𝑦 and 𝑧. Let’s begin by equating three 𝑦 to six. Three 𝑦 equals six. And so we’ll solve this equation for 𝑦 by dividing both sides by three. Three 𝑦 divided by three is 𝑦, and six divided by three is two. So we can say that 𝑦 must be equal to two.

Similarly, we can equate the other two parts of the equation. We can say that two 𝑧 is equal to six. And then to solve for 𝑧, we simply divide through by two. Two 𝑧 divided by two is 𝑧, and six divided by two is three. So we found that 𝑦 is equal to two and 𝑧 is equal to three. That’s not enough, though. We needed to find the value of 𝑦 plus 𝑧. Well, we can now say that that must be equal to two plus three, which is of course equal to five. And so, given the restrictions on our equation, we can say that 𝑦 plus 𝑧 must be equal to five.

We’re now going to consider how to find the solution set of an exponential equation when the exponents are binomials.

Find the value of π‘₯ for which eight to the power of π‘₯ plus two equals two to the power of π‘₯ plus four. Give your answer to the nearest tenth.

Here we have an equation that’s made up of two exponentials, but where the exponentials are binomials; they’ve got two terms in them. And in fact, the key to answering this problem is in spotting that each of the bases of our equation β€” that’s the big number, so eight and two β€” can be written as a power of some common number. In fact, if we recall that two cubed is equal to eight, we can rewrite the base on the left-hand side as two cubed. So, let’s see what that looks like. We get eight to the power of π‘₯ plus two equals two cubed to the power of π‘₯ plus two. And that’s actually really useful because we know that we can multiply the exponents now.

We recall that π‘₯ to the power of π‘Ž to the power of 𝑏 is π‘₯ to the power of π‘Ž times 𝑏. And so this becomes two to the power of three times π‘₯ plus two. Let’s replace this expression in our earlier equation. When we do, we find that two to the power of three times π‘₯ plus two is equal to two to the power of π‘₯ plus four. So, how does this help? Well, now that the base is the same on both sides β€” it’s two β€” we can say that for the equation to make sense, the exponents themselves must also be the same. That is, three times π‘₯ plus two is equal to π‘₯ plus four.

To solve this equation for π‘₯, we’re going to begin by distributing the parentheses on the left-hand side. And we do so by making sure that we multiply three by the π‘₯ and then three by the two. Three times π‘₯ is three π‘₯, and three times two is six. So, our equation is three π‘₯ plus six equals π‘₯ plus four. Since we’re looking to isolate the π‘₯, we’re going to begin by subtracting the smaller number of π‘₯. So we’re going to subtract π‘₯ from both sides. Three π‘₯ minus π‘₯ is two π‘₯, and π‘₯ minus π‘₯ is zero. So, our equation becomes two π‘₯ plus six equals four. Next, we subtract six from both sides. Two π‘₯ plus six minus six is two π‘₯, and four minus six is negative two.

Our final step to solve for π‘₯ is to divide through by two. And when we do, we find that π‘₯ is equal to negative one. Now of course, negative one is an integer solution, so we don’t need to do any rounding despite the fact that it tells us to in the question. What we can do, though, is check our solution is definitely right by substituting it back into the original equation. Eight to the power of π‘₯ plus two becomes eight to the power of negative one plus two. That’s eight to the power of one, which is eight. Then, two to the power of π‘₯ plus four becomes two to the power of negative one plus four. That’s two cubed, which is also equal to eight. And since these expressions are equal, we can say our answer is correct. π‘₯ is equal to negative one.

In our next example, we’ll look at how to solve an exponential equation that involves some absolute value.

Find the solution set of two to the power of the absolute value of eight π‘₯ minus 12 equals eight to the power of four π‘₯ minus four.

The key to solving this equation is to spot that we can rewrite eight as two cubed, thereby creating an equation where the bases are equal. With the expression we have on the right-hand side, if we replace eight with two cubed, we get two cubed to the power of four π‘₯ minus four. But then, of course, one of our rules for working with exponents can help us to simplify this. We know π‘₯ to the power of π‘Ž to the power of 𝑏 is the same as π‘₯ of the power of π‘Ž times 𝑏. When we’re dealing with parentheses, we can multiply the exponents, and so we can rewrite this as two to the power of three times four π‘₯ minus four.

And so then, our original equation can be rewritten as two to the power of the absolute value of eight π‘₯ minus 12 equals two to the power of three times four π‘₯ minus four. Now, this is really useful because now that the bases are the same β€” it’s two on both sides β€” we can say that for this equation to make sense, for everything to be equal, their exponents must themselves be equal. That is, the absolute value of eight π‘₯ minus 12 must be equal to three times four π‘₯ minus four.

Let’s distribute these parentheses by multiplying each term inside the parentheses by the three on the outside. And when we do, we get 12π‘₯ minus 12. So, how do we solve an absolute-value equation? Well, we know that the absolute-value symbol takes any negative input and essentially makes it positive. And so, what we do is we change the sign of what our absolute value is equal to. So we say that either eight π‘₯ minus 12 is equal to 12π‘₯ minus 12 or eight π‘₯ minus 12 is equal to the negative of 12π‘₯ minus 12.

Now, of course, if we distribute the parentheses on the right-hand side, we get negative 12π‘₯ plus 12. So let’s solve both equations. With our first equation, we’re going to subtract eight π‘₯ from both sides, and we get negative 12 equals four π‘₯ minus 12. Then we add 12 to both sides. And when we do, we get zero is equal to four π‘₯. Well, the only way for this to be the case is if π‘₯ itself is equal to zero. So, that’s one potential solution. Let’s solve the second equation. With our second equation, we’re actually going to begin by adding 12π‘₯ to both sides, so we get 20π‘₯ minus 12 equals 12. Then we add 12 to both sides, and we get 20π‘₯ equals 24. And then we divide through by 20, and we find π‘₯ is equal to 24 over 20, which simplifies to six-fifths.

What we’re going to do to check both of these solutions is to substitute them back into our original equation and check they both actually work. If we substitute zero into our original equation, the left-hand side becomes two to the power of the absolute value of negative 12 and the right becomes eight to the power of negative four. Then the absolute value of negative 12 is simply 12, and eight to the power of negative four is one over eight to the power of four. Now, if we actually evaluate these, we find two to the twelfth power is 4096. And then the right-hand side becomes one over 4096. Now, we can actually see these are not equal to one another. And so, actually, the solution π‘₯ equals zero doesn’t actually work. So we’re going to disregard that solution.

Let’s repeat this process with the second value of π‘₯. On the left-hand side, we get two to the absolute value of negative 2.4, and that should be equal to eight to the power of 0.8. But of course the absolute value of negative 2.4 is just 2.4. So let’s evaluate both sides. When we do, we find that they are both equal to roughly 5.27, so this solution does work. Now, of course, another way to check this would have been to rewrite two to the power of 2.4 as two cubed to the power of 0.8. And that’s because three times 0.8 is 2.4. Then we know that two cubed is eight. So we’re actually saying eight to the power of 0.8 is equal to eight to the power of 0.8.

Now, of course, the question was asking us to find the solution set to our equation. And when there’s only one value, we can use these sort of squiggly brackets to help us. The solution set is the set containing the element six-fifths.

In our final example, we’re going to look at how a bit of observation can help us to solve more complicated exponential equations.

Determine the solution set of π‘₯ to the power of π‘₯ squared minus 64 equals six to the power of π‘₯ squared minus 64.

Now let’s begin by noticing that the exponents on each side of our equation are in fact equal. And so, for both sides of our equation to be equal, one value of π‘₯ would be when the bases, the big numbers, are equal. In other words, if we have π‘₯ is equal to six, both sides of our equation are identical, so that is one solution. But are there any other options? Well, another way to ensure that two sides of our equation are equal is to have a power of zero since anything to the power of zero is one. And so what we could do is say that the exponent π‘₯ squared minus 64 is equal to zero.

Let’s solve for π‘₯ by adding 64 to both sides to give us π‘₯ squared equals 64. And then, finally, we take the square root of both sides, remembering, of course, to take both the positive and negative square root of 64. But since the square root of 64 is eight, we can say that the solutions to this equation are π‘₯ equals negative or positive eight. And so, so far, we have three possible values of π‘₯. But in fact, there is one more. And that solution is π‘₯ equals negative six. So, why does π‘₯ equals negative six work?

Well, imagine π‘₯ is equal to negative six. When it is, the exponent becomes negative six squared minus 64, which is negative 28. And we know that if π‘Ž is an even number, the negative π‘₯ to the power of π‘Ž is the same as π‘₯ to the power of π‘Ž. And so, the left-hand side is negative six to the power of negative 28. But because negative 28 is even, this is the same as six to the power of negative 28, which is what we get on the right-hand side. And so, π‘₯ equals negative six also has to be a solution. Now, of course, we only chose negative six because we wanted to match the bases. We wouldn’t go ahead and choose any value of π‘₯ that would make the exponents equal. It only works for negative six.

And so we can say the solution set to our equation is the set containing the elements six, negative six, eight, and negative eight.

In this video, we learned that we can use the rules of exponents to help us solve exponential equations. We saw that if we can create a common base by using these rules of exponents, we can then equate the exponents themselves and solve as normal. We saw that we should always check whether all answers given are actually valid by substituting them back into the original equation and that if we’re working with negative bases, then we should watch out for any possible even exponents that would create extra solutions.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.