Equations lie at the core of mathematics, and many equations involve exponentials, logarithms, and trigonometric functions. We’re going to discuss solution techniques for such equations mainly through examples. We’re going to build from simple examples to more complex ones, and we’re going to begin with exponential and logarithm. As we’ll see, there’s no unique solution technique, but it all depends on really knowing the properties of the functions involved, for example that exponential and logarithm are inverse functions, one to the other. Here’s the first example. We want to solve the equation for x, e to the 3 x equals 5. As you can see, the 3 x is up in the exponent here.
That induces us to take the logarithm of both sides, the natural logarithm, ln. When we do that, on the left we get 3 x because ln of e to the something is the something. That allows us to solve for x and we get the answer. The second example. We want to solve ln of x cubed equals minus 1. There’s more than one way to do it. Here’s a somewhat indirect approach. You bring the minus sign on the left side and you bring the minus sign into the logarithm with the reciprocal law for logarithms and you get that ln of 1 over x cubed is 1. Now what number is it whose natural logarithm is 1? There’s only one.
It’s e, so 1 over x cubed must be e, and from that, you find x, e to the power minus 1/3. A more direct approach that another student might have taken would be to directly take the powers of e on both sides, and on the left then you get x cubed, and on the right, you get e to the minus 1, and you still find, of course, the same value of x. Notice here, by the way, that there was a domain issue even though we didn’t mention it. Since the logarithm is only defined for positive x’s, the domain of this equation was x greater than 0. Well, our answers satisfied that, so there’s no problem.
Sometimes we have to think of the domain though a little more explicitly even from the start. So what are the facts we need to solve these types of equations? Well, they’re all the facts that you’re familiar with about exponentials and logarithms. It looks like a fairly long list, but let’s remember that the e of ln brings you back to x and vise versa. You have the same sort of fact for logarithm to another base other than e, log to base a. You have a bunch of rules about how logarithms work, logarithm of the product is the sum of the logs and so forth.
All these rules can be used, and in addition, it’s important to bear in mind the domain of the functions, the range, the monotonicity, and certain special values. It seems like a lot, but in fact, I think we already know most of these things. Let’s look at this example. We want to solve this equation. There doesn’t seem to be a domain issue here. If we take ln of both sides and on the left we get what you see. And how have I simplified that? Well, on the left-hand side, I use the rule for the logarithm of a power. And on the right-hand side, I use the fact that ln 1 is 0.
Now next, I deduce from what I attained that x is equal to minus 1. Why? Because I use the fact that ln 4 can’t be 0, and therefore, x plus 1 must be 0. So the answer is x equals minus 1. Another example, here’s an equation involving logs. Now with these equations, you’d better be aware of the domain right from the beginning. We notice that x has to be strictly positive for the two different logarithms to be defined. So that’s the domain of the equation. How are we going to solve it? Well, we’d like to bring everything in terms of the log with respect to the same base. So we’re going to work on log 27 of x.
There’s a rule that expresses that in terms of logs to base 3. I’ve used that rule here, and I’ve also used the fact that the log of 27 to base 3 is 3. Why? Because 3 to the power of 3 is 27.
Now once I’ve made the simplification of the log to base 27, I insert it into my original equation and I find easily that the log of x to base 3 must be 3. And now I deduce from that taking the powers of 3 that x is equal to 3 cubed or 27. That’s the answer, and notice that the answer is admissible. That is, it lies in the domain x greater than 0, as it must. Another example, a little more complex, this one, because as you can see, x is now present on both sides of the equation. We proceed to take ln of both sides, and after that, we have x as a linear factor on both sides.
We put the x’s on the same side of the equation as usual, and we see that x is multiplied by a certain number and it’s equal to another number, and we divide across and we find x. Nothing too hard. Now this equation is more complicated because the x occurs not only four times, but in a different appearing way, and it needs a little judicious factoring before we can solve this equation. You look for something that you can factor out on each side so as to reduce the number of x’s so to speak. Well, on the left-hand side, you notice that a factor of 3 to the x can be pulled out from both terms.
And on the right-hand side with a little thought, you see that the factor 8 to the x is common to both terms on the right. When you do this factoring, you then have reduced your original equation to one in which you only have the presence of x once on the left and once on the right. Now you can proceed as we did in the preceding exercises. That is you take ln of both sides having simplified a bit on both left and right, and having applied ln, you get a simple-looking equation, which you solve for x, and that’s the answer. Now more complex exponential equations will involve a certain function of a to the x equals 0.
And one of the techniques that very often is useful for these more complicated equations is a change of variables. We’ll see how this works, for example, in this equation. This is harder than any of the others we’ve seen so far because, not only is the x present on the left and the right, but there’s nothing you can easily factor out to have a multiplicative form as we did in the preceding exercise. So what do you do? Well, you look for some expression involving x that lends itself to a change of variable. Now with a little practice, you can see here that the term that suggests itself is 2 to the 2 x.
So you call that another variable, let’s say y. When you make that substitution into the equation, you get something, you see, which reduces to a standard quadratic equation for y. Well, we know how to solve those, the discriminant and all that, and you get the two possible roots of this quadratic equation. They are y equals 1 plus or minus root 2. Now we go back to the original variable. We remember that y is equal 2 to the 2 x or equivalently 4 to the x. So 4 to the x would have to equal 1 plus or minus root 2. But 1 minus root 2 is a negative number, and 4 to the x can’t be negative.
So we retain only the other possibility, 4 to the x equals 1 plus root 2. This is a simple equation like the very first one we solved a few minutes ago, and you wind up getting the answer for x, which can be expressed either in the first form you see or as the log to base 4 of 1 plus root 2, the same answer. Either answer is acceptable, of course. It’s just important to be able to recognize that they are the same number. Here’s our next example. Now the very form of this equation should lead you to be sensitive to the domain. Why?
Because you have an ln involved, so x will have to be strictly positive, and also you have quotients involved, and the denominator must not be 0. So the natural domain here is to say that x must be positive. The ln of x must be non-zero and also ln x plus 4 must be non-zero. So that gives you the set of all positive numbers except for two of them, namely 1 and e to the minus 4. That’s the domain. Now how are we going to solve the equation? Well, once again with a little experience, there’s a change of variables here that suggests itself. The appropriate change of variables is to set y equal to ln x.
When you plug that into the equation, it becomes an equation for y, which is equivalent to a quadratic equation for y whose roots are easily found, 3 and minus 2. And that tells you then that ln x would have to be minus 3 or 2. Both of those are possible. That gives you x equals e to the minus 3 or x equals e squared. Now you checked the domain, and you see that both these points are acceptable. They’re both in the domain, and therefore, a solution set consists of these two points. Here’s another example in which it should be clear that we need to be concerned about the domain right from the start.
In order for this equation to make sense, x has to be greater than 1. 2 x minus 1 has to be greater than 0, and x has to be greater than minus 1 for the three lns to be defined. That turns out to be equivalent, those three simultaneous conditions, to x being greater than 1. So the domain of this equation is the open interval 1 plus infinity. How are we going to solve it? Well, I’m going to work on the left-hand side to write it as the ln of a single number.
I use the well-known ln of a product rule to do that, and on the right-hand side, I bring the 2 that’s initially in front of the ln in as a power on the inside. So our equation then becomes ln of something equals ln of something else. Since ln is an injective function, it’s clear then that I have to have the two arguments being equal. That leads to a quadratic equation for x, which I solve without any difficulty, and I find x equals 0 or 5. Now, it would be a mistake to stop here because you have to see whether these points are in the domain of the original equation. As we see, only one of them is, x equals 5.
So we retain that one and reject the other, and our solution set is a singleton, the point 5.
The moral of this example, be sensitive to the domain of the equation right from the start.