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Contents

- 1 Equation Solving
- 2 Review Roots
- 3 Review Roots
- 4 Exponential Equation
- 5 Exponential Equation
- 6 Exponentital Equations Writing Expressions with the Same Base
- 7 Exponentital Equations Writing Expressions with the Same Base
- 8 Exponential Equations Fractions for Bases
- 9 Exponential Equations Fractions for Bases
- 10 Exponential Equations the Reciprocal of a Base
- 11 Exponential Equations the Reciprocal of a Base
- 12 Exponential Equations Practice 1
- 13 Exponential Equations Practice 1
- 14 Exponential Equations Practice 2
- 15 Exponential Equations Practice 2
- 16 Exponential Equations Practice 3
- 17 Exponential Equations Practice 3
- 18 Exponential Equations Practice 4
- 19 Exponential Equations Practice 4
- 20 Exponential Equations Practice 5
- 21 Exponential Equations Practice 5

When it comes to solving equations, we've learned a lot of different techniques to solve the quadratic ones. We saw the square root method, where we could take the square root of both sides of the equation. We used plus or minus to indicate the positive and negative root for this side. So here, x could equal positive or negative 3. We also could set the equation equal to 0 by subtracting 9 from both sides. This led us to factor. We could factor x squared minus 9 as the difference of 2 squares. So we have x minus 3 times x plus 3. Setting each factor equal to 0, we would get x could equal positive 3 or negative 3. So the square root method and factoring were two of the tools in our arsenal to solve this type of equation. We could even use the quadratic formula. If we subtract 9 from both sides. We'll have x squared minus 9 equals value of negative 9. This evaluates to plus or minus 3. Notice that any way we solve the original equation, we still get the same answers. Plus or minus 3 each time.

But, how would you solve an equation like x cubed equals 27? What do you think x equals in this case? Type it here.

You probably figured out that x must equal 3. And you know this based on the fact that 3 to the third power equals 27. So x must be 3. Good thinking if you got three.

So, from the last problem, we were able to figure out that x equals 3. But what if the exponent is not known? What if we had this equation? This second equation is called an exponential equation. It's called that since the exponent is the variable, and it's unknown. But we can apply similar reasoning. Let's take 27 and write it with a base of 3. We can rewrite 27 as 3 to the third power. And since these two bases are the same. We know the exponents must be equal in order for the two numbers to be the same. So here we know that this x must equal this exponent. X must equal 3. In general, whenever we have x to the a, equal x to the b, we know that the exponents a and b must be equal, since their bases are equal. This works in general so long as x, which is the base, is not equal to 0 or 1. So for an equation like this, what do you think x equals? Write it here.

For this exponential equation, the value for x equals 3. Great thinking if you found this number. We'll take 125 and write it with a base of 5. So we'll have these two bases are identical, we can set the exponents equal to one another. So x must equal 3.

But what can we do for an equation like this? Notice that these two bases are different. We have 16 raised to the x power. And 4 raised to the 3rd power. The and get the bases to match. We want them to be equal. So, we can rewrite 16 with a base of 4. We'll have 4 squared which is 16 raised to the x. And the right hand side will stay the same. We know by the properties of exponents when we raise a power to a power, we multiply these powers together. This means we'll have four raised to the 2 x, equal to 4 cubed. Now, since both of the bases are equal, we can set 2x equal to 3. These exponents must be equal to one another. And finally, we divide both sides by 2 to get one x is equal to three halves. Now, this value for x should make sense in our original equation. We'll have 16 raised to the three halves here. And we'll have 4 cubed on the right hand side of our equation. We know that a fractional exponent really represents a root. So, this really says that we have the square root of 16 raised to the equal to 4 cubed, so yes, this checks. You try solving this equation. Try rewriting these bases so that way they're the same, and then see if you can solve for x.

Here, x would equal 4 3rds. Nice thinking if you got this number. Now, if this gave you some trouble, that's okay. We haven't had a lot of practice rewriting numbers with different bases, so let's see how it works. I know that both 27 and 81 are the result of raising 3 to a certain power. For example, 3 cubed is a 3 here, and a 3 here. But before I do anything, let's combine these two exponents. When we raise a power to a power, you multiply these two together. This is how we get 3 to the 3x equals 3 to the fourth power. Now, since these two bases match, we set these exponents equal to one another. So we'll have 3x set equal to 4 here, and then we'll divide both sides by 3 to get x equals 4 that 27 raised to the 4 3rds equals 81. We know this is true since the cube root of 27 equals 3. And 3 to the fourth power is 81. So yes, our answer checks and we can be sure that it's correct.

Let's try another equation. This time, the bases are fraction. Now, don't be worried, it works the same way as before. We want to think about rewriting the right-hand side of the equation, so that way we have the same base of 3 5ths, as we do on the left. Once we have identical bases, of 3 5ths, we can set these two exponents equal to one another. So, try your best and see if you can find x.

Here, x would equal positive 3. Great thinking if you found this. Now, the easiest thing to do is to think about writing this with a base of 3, and this with a base of 5. 27 is the same thing as 3 cubed. And 125 is the same as 5 cubed. Now that we have 3 cubed divided by 5 cubed, I can pull out a cube from the numerator and denominator and just raise 3 5ths to the 3rd power. Remember that when you raise a fraction to an exponent, you can apply the exponent to each part of the fraction. So this is an equivalent statement, for this. Now notice, we have the same base as 3 5ths, so we set the exponents equal. x equals 3.

Try solving this equation. What do you think is the value for x here? You'll need to remember something about flipping bases. There's a certain type of exponent, or a value of an exponent, that can flip or do the reciprocal of a base. You'll use that to solve this problem.

Here x equals negative 3. Excellent work if you found this number. This one was tough. Now, the first thing to notice is that 125 is 5 raised to some exponent and 8 is 2 raised to some exponent. So let's rewrite this numbers with bases of can pull out this third power to have 5 halves raised to the third. The left-hand side of the equation just stays the same. And notice we're so close for these bases to be equal. We have 2 5ths on the left, but 5 halves on the right. So, let's flip this base over. Let's do the reciprocal of it. Remember that to flip over a base, or to make 5 halves become 2 5ths, we raise it to a power of negative one. So, this exponent of 3 becomes negative 3. Now we have identical bases of two-fifths, so x must equal negative three. The key idea to solving this problem is to remember to flip over or take the reciprocal of a base. To do that, we need a negative exponent.

now that you seen a variety of problems, let's get into practice. What do you think the solution for this equation? write that answer here

For this equation, x would equal 3 halves. Great thinking if you found this fraction. Now we want to start by rewriting these with the same base. So I know I'll multiply these two exponents to get 3 to the 2 x equal to 3 cubed. Now that our bases are identical, then we can set these exponents equal. So we'll have 2x equal to 3. Dividing both sides of the equation by 2, we'll have x is equal to 3 halves, or 1 and a half.

How about this equation? What do you think is x?

The solution for this equation is 5 halves, or two and a half. Nice work if you got this one right. Again, we want to use the key idea of rewriting these with the same base. I know 4 is 2 squared and I know 32 is 2 raised to the 5th. So, now I'm working towards having the same base. I want to combine these exponents by multiplying. So, I'll have 2 to the 2x equals 2 to the 5th. I have 2 raised to the 2x and 2 raised to the 5th. These must be equal, so I know that the exponents have to be equal, since the bases are the same. So, we'll have 2x set equal to 5. dividing both sides by 2 we find x is equal to 5 by 2

How about this equation. What do you think is x?

W'll start by rewriting 9 16ths as 3 squared divided by 4 squared. We can do this since we know 3 squared is 9 and 4 squared is 16. Next, we're going to pull out our exponent of 2 from each part of the fraction. So we'll just have 3 match. I have 4 3rds here, but I have 3 4ths here. In order to get these two bases to match, we need to flip this base. So, we'll use a negative exponent to do that. So, 3 4ths squared is the same thing as 4 3rds raised to the negative second power. Remember, this negative 1, or this exponent of negative 1, flips over our base. Now we have two bases that are the same, so we can set the exponents equal. X must equal negative 2.

Try this problem, this might be the hardest one yet. What do you think x is?

In this case, x must equal negative 4. Great work if you found this number. We start by taking our fraction of 1 divided by 81, and we rewrite it as 81 raised to the negative first power. Remember that this negative 1 will flip over our base. This would make us have our reciprocal of one divided by 81. So we know these two are equal. Next, we want to make these bases both be 3. So, I rewrite together to get 3 to the negative 4th. Now my bases match so the exponents must match as well. x equals negative 4.

Here's our fifth and final problem for exponential equations. We have 16 raised to the 2x plus 1, equal to 64 raised to the x plus 3. Notice that 2x plus 1 is our entire exponent, and x plus 3 is our entire exponent. What do you think x equals in this case?

When solving this problem, x equals positive 7. Now, that's great work if you got this one correct. Now, there were two ways to do this problem, we could write these two bases with a base of four or we could write these two bases with a base of two. I'll show you both methods, but first let's do the one with a base of four. I know 16 is the same thing as 4 squared. And I also know 64 is the same thing as 4 cubed. So we write each of these bases with a base of 4, raised to a certain power. These other exponents stay on the outside of the parenthesis. Now, we have a power raised to another power. Which means we can multiply these two powers together. The same is true on the right hand side. We have 4 to the 3rd raised to x plus 3. So, I'll multiply this power by this power here, and I'll multiply this power of 3 by the power x plus 3. Notice too that I'm putting the power in parenthesis. It's because I need to distribute this 2 to both of these terms. The same is true for the 3. We multiply the 3 by x, and the 3 by 3. Distributing the 2 here and here, and distributing the 3 here and here, we get these powers. Now we have bases of 4, and we can set these two exponents equal to one another. So, we'll have 4x plus 2, set equal to 3x plus 9. Next, we'll subtract 3x from both sides of the equation to get 1x plus 2 equals 9. Then, we'll subtract 2 from both sides of the equation to get our answer of X equal to 7. Now the other way to do this problem is to rewrite the base of 16 and the base of 64 with powers of 2. 16 is equal to 2 to the 4th and 64 is equal to 2 to the 6th. Next, we have powers raised to powers on both sides of the equation. So we can multiply these two exponents together, here, and we can multiply these two exponents together here. Now we have 2 raised to the 8x plus 4, distributing this 4. And we have 2 raised to the 6x plus 18, distributing this 6. Now that both bases are identical, we set these exponents equal to one another. So we'll have 8x plus 4 equal to 6x plus 18. We'll subtract 6x from both sides of the equation to get 2x plus 4 equals 18. Next, we'll subtract 4 from both sides to isolate the 2x. So, 2x will equal 14. Finally, we divide both sides by two to get 1x equals seven, our solution.