Inequalities with logarithms and exponentials

We’re now going to look at inequalities and, to begin with, inequalities involving exponentials and logarithms. We’re going to see that solving these inequalities is a lot like solving equations with exponentials and logs, mainly because these functions have such nice monotonicity properties. And of course, all the properties of exponentials and logs will play a role, in principle. We’re going to look at examples, beginning with this one. We want to find the x’s for which this inequality is satisfied.

Now, if it were an equation, we’d take the ln of both sides, and we’re going to do the same thing here, and the reason that we can get an equivalent inequality after taking the natural logarithm is because ln is a strictly increasing function, so it preserves the inequality, so to speak. Of course, the numbers involved were also positive, so ln was defined at these numbers.

When we now simplify, having taken ln, we get a simple inequality in x, we express the same inequality but with the x’s all on one side, we factor out the coefficient of x, which turns out to be 3 ln 5 minus 2 ln 4, and I would like to simply divide across by that to express the answer. But careful, when we divide across, the inequality– the sense of the inequality is only preserved if we’ve divided across by a positive number. So have we divided across by a positive number? Well yes, because we can check that 3 ln 5 minus 2 ln 4 is positive.

If we put it all together as a single logarithm, using the familiar laws of logarithms, we see that it’s the logarithm of 5 cubed over 4 squared. Well, 5 cubed over 4 squared is greater than 1, therefore its ln is strictly positive. And so our answer is justified and we found it. Let’s look at another example. There were, by the way, in the previous exercise, no real domain issues, but now in this type of inequality, we see that there is a domain issue because there’s a square root present on the left-hand side, and therefore the domain D of this inequality is going to consist of those x’s for which 3 minus c to the x is greater or equal 0.

That is, x has to be no bigger than ln 3. In that domain we can square both sides, simplify the equation a little bit, and you wind up getting what you see, which– whose solution set is not obvious by any means, but which evidently tempts us to make a certain change of variables to solve this new inequality we’ve generated. The evident choice, we define a new variable t by setting it equal e to the x. In terms of t, our inequality is just a quadratic inequality, so we want to know the t’s for which this quadratic function, t minus 2 times t plus 1, is greater than or equal 0. That’s easy to solve, graphically or otherwise.

We find that the t’s must either be to the left of the smaller root, minus 1, or to the right of the bigger root, 2. This then identifies two cases. Recalling that t is e to the x, either e to the x is less than or equal minus 1, but that’s crazy, e to the x can’t be negative, or e to the x will be greater or equal 2, and that gives us x greater or equal to ln 2. It would be a mistake to stop here because we need to retain only these values of x which are consistent with the original domain of the inequality.

When you look at that domain, that tells you that the x’s you should keep as your solution set are those in the interval, the closed interval ln 2, ln 3.

A remark. We have just solved the inequality that you see and we found the solution set S. What if we wanted to solve the opposite inequality, that is, the square root greater than 1 minus e to the x? You might be tempted to think, well, it’s the other values of x, that is, the complement of the set S. But that’s a mistake because there was a domain for the original inequality which is the same as for the new one, and the set of solutions of our new reversed inequality will be, yes, the complement of the set S, but relative to the domain D. That is, the solution set for the new inequality is D set minus S.

We take D and we take away the points that are in S and so we get the open interval minus infinity to ln 2.