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Contents

- 1 Notes for Lesson 33: Solving Exponential and Logarithmic Equations
- 1.1 Bacteria Bread Money Earthquakes
- 1.2 Bacteria Bread Money Earthquakes
- 1.3 Solve an Exponential Equation
- 1.4 Solve an Exponential Equation
- 1.5 One to One and Inverse Properties
- 1.6 One to One and Inverse Properties
- 1.7 Tricky Base
- 1.8 Tricky Base
- 1.9 Prepare to Calculate
- 1.10 Prepare to Calculate
- 1.11 Change the Base
- 1.12 Change the Base
- 1.13 Step One
- 1.14 Step One
- 1.15 Pick a Base and Solve
- 1.16 Pick a Base and Solve
- 1.17 Find x
- 1.18 Find x
- 1.19 Pretend Variable
- 1.20 Pretend Variable
- 1.21 Log Base 2
- 1.22 Log Base 2
- 1.23 Approximate t
- 1.24 Approximate t
- 1.25 Alternative Quadratic
- 1.26 Alternative Quadratic
- 1.27 Equation in Terms of u
- 1.28 Equation in Terms of u
- 1.29 From u to x
- 1.30 From u to x
- 1.31 Back to Banking
- 1.32 Back to Banking
- 1.33 Grant Saves Money
- 1.34 Grant Saves Money
- 1.35 Hurricanes, Earthquakes, and Hearing Tests
- 1.36 Hurricanes, Earthquakes, and Hearing Tests
- 1.37 Shift It Up
- 1.38 Shift It Up
- 1.39 Big Number
- 1.40 Big Number
- 1.41 Isolate 2x+3
- 1.42 Isolate 2x+3
- 1.43 The Search for x
- 1.44 The Search for x
- 1.45 Isolate lnx
- 1.46 Isolate lnx
- 1.47 Solving for x
- 1.48 Solving for x
- 1.49 Just One Log
- 1.50 Just One Log
- 1.51 What equals 0
- 1.52 What equals 0
- 1.53 A Quadratic Equation
- 1.54 A Quadratic Equation
- 1.55 Actual Solutions
- 1.56 Actual Solutions
- 1.57 Earthquakes in Japan and California
- 1.58 Earthquakes in Japan and California
- 1.59 Splitting Logs
- 1.60 Splitting Logs
- 1.61 Subtract the Equations
- 1.62 Subtract the Equations
- 1.63 A Single Log
- 1.64 A Single Log
- 1.65 Intensity Ratio
- 1.66 Intensity Ratio

In the last few lessons, you've learned a lot of new information about exponential and logarithmic functions. You've seen their graphs. You've learned about properties that apply to them, and you've also learned how to evaluate and manipulate them. In the coming videos, we're going to explore various strategies to solve a variety of exponential and logarithmic equations. A few lessons ago, we touched on an application problem where we considered the amount of bacteria which remained alive after a patient was medicated. This lesson is going to give you the tools you need to solve other application problems as well. Maybe you're earning interest on a certain amount of money that you've put in a savings account. Then we'd have an equation like this. of bread, but you're about to go on vacation. We could use an equation like this to figure out how much mold would grow on the bread after a certain amount of time. Or, what if you've just been through an earthquake, and you want to figure out how intense the quake was? Well, you could use equations like these two to talk about that. This one characterizes the earthquake that happened in Japan in 2011. And this one talks about the worst earthquake that ever happened to California, which was in 1857. These scenarios are all going to be super interesting to look at, but I think we should wait a bit before we get to them. For now though, let's stick with something pretty simple. You've been learning with exponents a lot lately, so I think this equation is a good choice. We have do this in your head or if that seems tricky you can try guessing and checking using a calculator.

We can rewrite 64 as 4 to the third power, which means that we have 4 to the x equals 4 to the 3. That means that x equals 3.

Solving this equation, might have felt really simple to you but there's actually something really important going on here. This is highlighting a certain property that exponents have. If we have an equation like this one, a to the x equals a to the y, then we know right away that x must equal y. This property also works the other way around. If we know that x equals y, then we also know that a to the x is equal to a to the y. Now that we've talked about that concept, here's a slightly harder problem to try. This time we have 4 to the x minus 1 power equals 64. What is x now?

Once again, let's rewrite 64 as 4 cubed. We know that our exponents need to equal each other, so we have x minus 1 equals 3. Adding 1 to both sides gives us a final answer of x equals 4.

This is all well and good if our numbers work out nicely. But what if we don't have numbers that are as pretty as these? That work this easily together. Let's stick with 4 to the x. But how about on the right hand side of the equation, we put something different, like 121. Now, this problem is not as simple as the last 2 .We can't easily make 121 into a power of 4. So I think we need to go back to our knowledge of properties. In particular, properties of exponents and logs. We have some one-to-one properties. There's this one that we just talked about. If we know that A to the x equals A to the y, then we also know that x equals y. And similarly, we have this second one. If we know that log base A of x is equal to log base A o fy, then we also know that x equals y. We also have some inverse properties. Properties that demonstrate that exponential and logarithmic functions are each other's inverses. The first one is that a to the power of log base a of x is equal to x. Here's the first one. a to the power of log base a of x equals x. And the second one says that log base a of a to the x is equal to x. This is just showing that logs and exponents undo one another. Now, keeping these in mind, let's go back to that equation we were working with. And here it is again. 4 to the x equals 121. Now as we did in earlier lessons, I think we should take the log of both sides, after all we want to extract the x from the exponent here. One of those two inverse properties was this one, that if we have something like a to the x, just as we do here, if we take log base a of that we'll just end up with x on this side of the equation. That sounds pretty good to me. So thinking about the equation that we have, what base should the log that we apply to both sides be? In other words, what should a here equal?

We should use a log of base 4 here. Note that in this property right here, the base of the log is equal to the base of the exponent. Since our base on the left hand is 4, we want to use log base 4.

So, let's use it now that we've figured out what we want to do. Please take the log base 4 of both sides and then simplify the left-hand side using the inverse property.

The first step is to take the log base 4 of both sides. So we have log base 4 of 4 to the x equals log base 4 of 121. Then, we just simplify over here. Using our inverse property we talked about, log base 4 of 4 to the x is just x. Since this log undoes this exponent. That leaves us with x equals log base 4 of 121.

So we're in great shape. We have an expression for x. But are we really done yet? I don't think so. We know that what we have over here is actually just equal to a number, but what number is it? Since we can't rewrite 121 off the top of our heads, as a power of 4, we going to need to do a little bit of work here. What property of logs do you think we need to use to find out the value of log base 4 of 121? Should we use the product property, the quotient property, the power property, or the change of base property? Think about what's keeping us right now from evaluating this side of our equation.

We need to use the change of base property. All of these other properties would leave us with logs of base 4. So they're not going to enable us to use our calculator to come up with an answer.

Using the change of base formula and picking either log base 10 or natural log, we get a value for log based 4 of 121. What does that mean that x equal?

Since most calculators, including the one I'm going to use, have buttons for log base ten and natural log, I've shown you two ways that you can could pretty easily figure out how to evaluate log base four of 121. If we want to convert it to something involving log base ten, it would become log base 10 of 121 over log base 10 of 4. If instead, you wanted to use natural log, it would become natural log of 121 over natural log of 4. Plugging this into a calculator gives us the same thing for each one. Rounding, we get about 3.459 for both. That means that x is about equal to 3.459. Awesome. We solved an exponential equation.

Do you remember solving linear equations? Well, we're going to use some of the same techniques we learned all those lessons ago to change equations like this one, so that we can use our techniques using logs to find out what x is. It's pretty common in mathematics to try to change a problem to be in a form of what we've seen before. So here, we have the equation e to the x minus 3 equals 43. For right now, let's just think of e to the x as our variable. Now if we do the same thing to both sides of our equation, what is e to the x equal to?

In order to get e to the x by itself, we need to add 3 to both sides of the equation. On the right-hand side, that's going to give us 46.

So now our equation looks just like the ones we were solving earlier. There's an exponent term only on one side and a constant on the other side. It makes sense then to use the same technique we did before. Taking the log of both sides to get x by itself. Now, think about what base log we need to use to get x by itself. And then, think about if that log is something you can plug into a calculator. Once you do get an answer for x, please round to three decimal places.

We know from our identity properties that we want to use a log of the same base as we see for this exponent here. A log of base e is the natural log. Taking the natural log of both sides gives us ln of e to the x equals ln of 46. Now, conveniently, ln is the button on the calculator. So let's plug this in and see what number we get. Any natural log of 46. And here is our long number. Remember, even this decimal is rounded. But we're only going to look at the first three decimal places. You know, the left hand side of our equation is just going to equal x and that decimal we got rounds to 3.829. This should really be an approximately equal to sign. Awesome. We solved a more complicated exponential equation.

Here's another example where we can use out techniques from solving linear equations. Again, I'd like you to start out by thinking of e to the x as our variable. Once you get it by itself on one side of the equation, you can use logs to solve for x. Once you find an answer for x, please round to two decimal places.

Let's think first how to get e to the x by itself. Before we do anything else, we need to subtract 23 from both sides. That leaves us with 3e to the x equals don't think it's going to be very pretty. So, I'm just going to leave this as an improper fraction for now. So we have e to the x equals 323 over 3. Now, I want to get x by itself, and what do I need to do? Well, I need to take the log of both sides and the base of that log needs to be the same as the base we have here, which is e. So I have ln of e to the x equals ln of 323 over 3. We know that the left-hand side will reduce just to x. That's why we picked a log of this base after all. To figure out what the right-hand side evaluates to, we have to use our calculator. Okay, so natural log of 323 divided by 3. Double check that we've add the correct thing here and that looks right to me. That equals 4.679 and so on. We said we wanted our answer to just two decimal places, so that should be 4.68.

Here's another equation for you to solve. And the ones we've been working with have definitely gotten gradually more complex, so this one is another step up. We have 3 times the quantity times 2 to 3t minus 2, plus 6 equals 15. Now you could approach this equation in the exact the same way as when we did just before this. You'll just have a few more steps at the end to reach the solution. Now, you might want to take a couple seconds right now and pause the video to try solving all the way to t on your own. If you feel like that might be a little bit much, let's go through it in a slightly more step-by-step manner. Now of course, there are many different ways to solve this problem, and I'm just going to show one of them. First, we're going to treat 2 to the 3t minus 2 as our variable. So, we write the equation isolating 2 to the 3t minus to work with, and sort of weird to think about as a variable, then you can substitute in another letter, like u for instance, in place of this whole thing. Then just solve for u.

First, we need to subtract 6 from both sides. Then, to get 2 to the 3t minus 2 by itself, we need to divide both sides by 3. That leaves us with 2 to the 3t minus 2 equals 3.

Since we have a base of 2 over here for our exponent, we need to use a log of base 2, since that's the inverse. Now we've done this before, so I'm just going to write it out for you. That gives us log base 2 of 2 to the 3t minus 2 equals log base 2 of 3. For now, I'd like us to leave the right-hand side as is. But what happens to the left-hand side if we simplify it? What belongs in this space?

Our log base 2 and 2 to a power cancel one another out. So all we're left with is what's in the exponent right here. That means the left-hands side should be

We've done a ton of work so far and we're super close to getting t by itself. So here's the equation we have right now. 3t minus 2 equals log base 2 of 3. I'd like you to solve this equation for t coming up with just a number that it's equal to. Please round the solution that you get to three decimal places. Now just a hint, in order to use your calculator to evaluate this you're going to need to change bases. So keep that in mind as you move forward.

To get t by itself we first need to add 2 to both sides. Then we just need to divide both sides by 3. So now we have t equals 1 third times log base 2 of 3 plus 2. Well what do we do with this? We know that we want to find out what number this is equal to. But our calculator doesn't have a button for log base two. That means that we need to convert this log base two of three into something that's either a natural log or a log base 10. I'm feeling in a natural log mood. So let's do that. Know that log base two of three will be equal to the natural log of three. Over the natural log of 2. So, let's substitute this into this spot. OK, so now we have t equals 1 third times the natural log of 3 over the natural log of 2, plus 2. This is calculator appropriate, so let's pull out the calculator. I am just going to enter all of this in. I have 1 third times the natural log of 3 divided by the natural log of 2. Plus 2. And here's what that equals. T is about equal to 1.195. Awesome, we went through a ton of steps to get to our answer, but we did arrive there eventually. And along the way you got to practice a ton of different techniques that you've learned throughout the entire course. Awesome job.

For the past few questions, we've looked at equations that we can treat as linear equations, up to a certain point, of course. But we spent a bunch of time in this course, also talking about quadratic equations. And in fact, we're going to need to use those techniques as well for certain equations that use exponents. Let's look at example of one of those right now. Here we have e to the 2x minus 5e to the x plus 6 equals 0. Now, it might not look like it right away, but this is actually a quadratic equation in e to the x. To see this more clearly, let's make a little substitution. Let's let u equal e to the x. How then, would you rewrite the first 2 terms of our equation in terms of u? Please fill in these pink spots with the proper numbers, variables, or expressions.

We know that e to the 2x is the same as e to the x squared. And since we're saying that u is equal to e to the x. This is just equal to u squared. For our second one, 5e to the x. That's just equal to 5u.

Using these substitutions you just came up with, please rewrite our equation here in terms of U. Then, find two solutions for U. You can use factoring to do this.

Substituting in e to the 2 x equals e squared, and 5 e to the x equals 5 u. We end up with u squared minus 5 u plus 6 equals 0. Now it's time to factor. This is equivalent to u minus 3 times u minus 2 equals 0. Then it's super easy to solve for u. We have u minus 3 equals 0 and u minus 2 equals 0. That gives us u equals 3 or u equals 2 as our solutions.

You might be thinking, great we solved our equation. But remember, we've only solved for u so far. What we really want to solve for is our original variable, x. Since u was actually equal to e to the x, what we really have here is e to the x equals 3 and e to the x equals 2. However, you know how to solve each of these equations. So, please solve for x and round each of your answers to three decimal places.

All I need to do to solve for x is take the natural log of each side of each equation. For e to the x equals 3, that gives us x equals ln 3, which is about equal to 1.099. For e to the x equals 2, we end up with x equals ln of 2, which is about to 0.693. Now let's remember what these mean. These are both zeros of the function f of x equals e to the 2x minus 5e to the x plus 6. Now in the practice section, you'll be able to see how this works a little bit more clearly.

To finish off the section on exponent methods. Lets look at an interest problem. Lets say that Grant puts $1000 in an account that receives 3% interest per year compounded yearly. How long is it going to take for him to have $1250 in the account? Well, let's remember that equation that we used for calculating compound interest before. It was A equals P times 1 plus r over n to the nt. Here A is equal to the money in the account. P is equal to the principal, the amount initially invested. R is the interest rate per year written as a decimal. N is the number of times interest is compounded each year. And t is equal to the number of years that have passed. So, what we really need to come up with right now is an equation for A after t years, of course, making it fit this situation. Please write that in here.

Plugging in the numbers from our problem, we know that P is equal to 1000, and R is equal to .03 N is equal to 1, which makes things pretty simple. Then into our equation for A is A equals 1000 times 1 plus 0.03 to the t.

We are told that the amount Grant wants in account eventually is $1250. So, lets substitute this in, in the place of A. That makes our equation 1250 equals variable, t. And conveniently, that's what we wanted to solve for anyway. So please solve for t using this equation. Once you have a number that t is equal to, please round up to the nearest integer or to the nearest year, since interest is paid into the account yearly. One more little hint for you. The base of a log doesn't have to be an integer. Think about that as you go through the problem.

To get t by itself you first want to divide both sides of the equation by 1000. On the left-hand side that leaves us with 1.25 and simplifying what we have in the parenthesis here. We have 1.03 to the t on the right-hand side. Now all we have over here term of an exponent. So we just need to take some sort of log of both sides to get t down out of the exponent, and by itself. Now as we've seen before the log we need to use should have the same base as this base right here, 1.03. So, interestingly enough, using log of base 1.03 for both sides. Well, that's certainly different from anything we've seen, but it works in the same way as a log with the base of any other number. On the right-hand side here, we just get t. And on the left-hand side we have this log expression that we would of course like to evaluate. Now I usually prefer having t on the left-hand side, so I'm just going to switch it over there now. That just makes things easier for me to see. Now since we don't have a log based 1.03 button on the calculator, we're going to need to use change of base. How about this time we use log of base 10. That means we're going to have, log base 10 of 1.25 divided by log base 10 of 1.03. Let's plug it in and see what that equals. So here's our famous calculator. Log of 1.25 divided by log of 1.03. What does that equal? Let's find out. 7.54914 and so on. Remember though, since Grant's interest gets compounded only once a year, we need to round this up to 8, the next highest integer. So, that is our answer. After 8, years Grant will have over $1250 in the account.

So far in this lesson, we've been working with equations involving exponents. So now let's look at the other side of the coin, equations involving Logs. Have you ever had a hearing test? Or maybe you felt an earthquake or seen a hurricane. All of these things are measured using logs. At the end of the section, we use the properties that you've learned about logs and the techniques that we're going to discuss. To compare the intensity of two different earthquakes. In the equations we're going to look at, the variable, which is oftentimes going to be x, will be part of the input to the log function. That means that to access is, we're going to need to use the inverse of whatever log function we have, since that will undo what the log has done to the x. So, what does that actually mean? Well, before, we were using each side of our equation, as the input to a log function, because we had exponents. But now, we're going to use each side as the input to the exponential function. So let's actually do some work with this equation. Ln of x equals 10. What base do you think will be best to use for our exponent, since of course, we want the inverse of the natural log?

We're going to want our base to be e, since natural log is a log of base e.

Use each side of the equation now as the exponent of e. And then simplify the left-hand side. What equation does that leave us with? Remember that inverse property we talked about before. A, taken to the power of log base a of x is just equal to x.

If we use each side as the input to an exponential function with e as the base. Then we have e to the lnx equals e to the 10. The left hand side, just simplifies to x, because of that inverse property that we talked about. And the right hand side, is just equal to e to the 10 still, since we haven't evaluated this yet.

We know that x is equal to e to the 10, but what number is that equal to? Use a calculator to evaluate this and round to the nearest integer.

Here's our calculator and all we need to do is plug in e to the power of 10. So remember, luckily we have a button for the number e, it is right here. That'll take us to the power, remember you press the button x to the y and our exponent is 10. So now we enter that. Great. E to the 10 equals wow a big number. 22026 and some decimal. So round it to the nearest integer, that's 22026. Awesome.

Let's take a look at another logarithmic equation. Here we have log base 2 of couple of steps. First things first, why don't you just try to get 2x plus 3 by itself. What belongs over here?

To make this input to our log by itself, what we need to do is take the inverse of log base two. That means we need to use an exponential function where two is the base. So we have 2 to the log base 2 of 2 x plus 3 equals 2 to the 4. Since the log base 2 is undone by this, all we have left on the left hand side is 2x plus 3 and on the right hand side, 2 to the 4 or 16.

Now all we have is a linear equation, awesome. Please solve for X.

Subtracting three from both sides and then dividing by two gives us X equals 13 halves or six and a half.

Now, something to notice about this equation we were just working with. Is that one thing that was convenient about it was that we only had a log on one side of the equation. Because of that, we are ready right away to use this as an input to an exponential function. What if that wasn't the case, though? Let's say we had something like this, 3 plus 4 lnx equals 6. We know that were going to need to isolate natural log here, so let's go ahead and start out by doing that. What belongs on the other side of the equation now?

Subtracting 3 and then dividing by 4 gives us natural log of x equals 3 over 4.

Now, I'd like you to actually solve for X. Please round your answer to three decimal places.

To get rid of our natural log here, we need to use lnx as the input to the exponential function with e as the base. So that gives us e to the lnx equals e to the 3 4ths. We know the left hand side just simplifies to x and the right hand side rounded to three decimal places is 2.117

Here's another equation for you to solve. Log base 10 of 5 x plus log base 10 of x minus 1 equals 2. To start off please condense the left hand side here to be the logarithm of just a single quantity. So in other words use the properties of logs that you learned to combine these two log expressions.

According to the product property, this is equal to log base 10 of 5x times x minus 1.

To keep moving forward with solving our equation, I'd like you to next to get rid of our log here. Then move all non 0 terms over to the left-hand side of the equation and simplify.

Using each side here as the exponent of 10 to some power. We end up with 5x times x minus 1 on the left and 10 squared on the right. That means that our equations is actually 5x squared minus 5x equals 100. Subtracting 100 from both sides gives us, 5x squared minus 5x minus 100 equals 0. I notice however, there's a common factor of 5 in all the terms on the left-hand side. So let's divide it out. That leaves us with x squared minus x minus 20 equals 0.

So the equation we're left with right now, is x squared minus x minus 20 equals

I've started out by factoring the left hands side. That gives us x minus 5 times x plus 4 equals 0. That means that x equals 5 or x equals negative 4. So, these should be the solutions to the original equation we started out with.

We've come up with, as solutions to the equation, log base 10 of 5x plus log base 10 of x minus 1 equals 2. Are x equals 5 and x equals negative 4. Something doesn't seem quite right here, though. What do you think would happen if we substituted in either these values of , in the spots of x in our equation. Would both of these inputs give values of our function that are allowed? Think carefully about the domain of the log function. So which solutions are correct?

The function f of x equals log based 10 of x can only have inputs that are positive. If we substitute in x equals negative 4, we'll find ourselves trying to take here the log of negative 20, or here the log of negative 5. We take here log base 10 of negative 20 and add to that log base 10 of negative 5. And yes, we get an error, just as we suspected. This isn't allowed because negative that x equals negative 4 is not a solution. Nothing goes wrong though for x equals 5. We get perfectly valid values when we plug in five in the spot of X here. That means that only X equals five is a solution.

So let's finally get to our earthquake problem. We're going to look at 2 earthquakes. One of them happened in Japan in 2011. This earthquake had a magnitude of 9.0. The other earthquake we're going to talk about happened in California. It was in 1857 and it had a magnitude of 7.9. This is the largest known earthquake in California. Now what do these numbers actually mean? Well, lets talk about the equation for the magnitude of an earthquake. Lets say that R is the magnitude of an earthquake on a Richter scale. Then for R we have the equation R equals log base 10 of I over S. Here, I is the intensity of the earthquake and S is equal to some constant. It's supposed to be the intensity of some sort of standard earthquake, which just has a fixed value. We don't need to know this value though, since we'll be eliminating it from our equations eventually. So here, once again, is our information about our two earthquakes. How does this information fall into an equation of this sort? Well, we know that R is the magnitude. So that means that over here for Japan we'll have 9.0 equals log base 10 of J over S. I'm just going to use the letter J instead of I, to show that this in the intensity of the Japanese earthquake. For California, we'll have something very similar. We're just going to instead have and R value of 7.9 and I'll use the letter C to represent the intensity of this quake, C for California. Now notice that in both of these log equations, we have a fraction as the input to the log. What we're going to want to do first is to split each of these up into two separate logs, since otherwise you're not going to be able to get rid of the S. So, which property do we need to use in order to do that? Should we use the product property, the quotient property, the power property or the change of base property.

We need to use the quotient property. Log of a over b equals log of a minus log of b. Both of these equations have quotients as the input to the logs.

Now that we know, we need to use the quotient property, let's do it. Please rewrite the right hand side of each of these equations, split into two separate logs.

Over here we end up with 9.0 equal log base 10 of J minus log base 10 of S. And here we have 7.9 equals log base 10 of C minus log base 10 of S.

I said before that we were going to want to compare our two earthquakes. But in order to do that we really need them in one equation together. To make that happen, I'm going to suggest that we subtract one of these equations from the other. Now we're allowed to do that because we have equal quantities lining up with one another. Lets see what happens. So please carry out the subtraction. Subtracting the entire California equation from the entire Japan equation. Now make sure that you let this negative sign distribute to every term we have here. What equation do we end up with?

side of the equation, we still keep our log base 10 of J, since nothing down here has cancelled that out. And we're subtracting log based 10 of C. Now, we have minus log based 10 of S here, and we're actually adding one down here. So, those two cancel out and this is all we're left with.

So now we have a nice equation. 1.1 equals log base 10 of J minus log base 10 of C. Now I'd like you to combine these two logs using the quotient property once again into a single log.

The quotient property tells us that this should just be log base ten of J over C.

Now that we have a single log equal to a constant, we can finally find j over c. This will tell us how many times greater the intensity of the earthquake in Japan was, than the one in California. So please solve for j over c. Round to one decimal place.

To get rid of our log here, we need to take both sides as the exponent of 10. The left hand side becomes 10 to the 1.1, and on the right hand side that will just leave us with J over C. Now let's use our calculator. 10 to the 1.1 is equal to 12.589 and so on, or rounded, about 12.6. So what does this mean? Well, that means that the intensity of the Japenese earthquake was about 12.6 times greater than the worst California recorded earthquake.