Before studying this post, make sure you have practiced the techniques taught in Exponential Expressions, Part 1. This section teaches you how to take the techniques you learned there, and bring them up to the next level.

## Multiples of 10 as Bases

Now that you've mastered 0^{0}through 10

^{10}, one of the easiest ways to go beyond just the exponents you've learned is to use bases that are multiples of 10. The same 10 trick that helped you learn how to take 10 to any exponential power will help here, too.

As a simple example, let's work out 20

^{3}. 2

^{3}is easy, that's 8. Since the exponent is 3, we add 3 zeroes after the 8, giving us a total of 8,000.

For any of the bases you've learned, adding a single 0 after it simply mean adding a number of zeroes equivalent to the exponent to get your answer. What is 90

^{5}? 5,904,900,000! Why? 9

^{5}, as you already know, is 59,049. The exponent of 5 tells us to add 5 zeroes after that, giving us 5,904,900,000.

Just this simple idea alone will help you deal take any multiples of 10, all the way up to 90, to the 10th power!

## Multiples of 100, 1000, and beyond

What if you needed to figure 500^{6}? The first step is simple, you just figure 5

^{6}, which you already know is 15,625. How do we handle the zeroes? You might think that the 6 just indicates adding 6 zeroes, giving us 15,625,000,000 (15.625 billion).

However, if you perform this problem with a calculator, you get 15,625,000,000,000,000 (15.625 quadrillion)! What went wrong? When working with a base ending in a single 0 (e.g., 40, 50, 60), the 6th power would properly indicate adding 6 zeroes. However, when your base ends in a pair of zeroes (e.g., 400, 500, 600), an exponent of 6 means you need to add on 6 pairs of zeroes. That's why the correct answer has 12 zeroes at the end, not 6.

This same basic idea holds true for any number of zeroes added to the end. 6,000

^{9}means you need to add 9 sets of 3 zeroes to the end of 6

^{9}. Are you ready for this? The answer to that is 10,077,696,000,000,000,000,000,000,000,000,000, or 10.077696 decillion! Did you ever think you'd be able to do a problem with an answer that large in your head?

## Presenting Multiples of 10

With the original problems up to 10^{10}, you've probably practiced them enough to be able to give the proper names of the large numbers. For example, when someone asks you 7

^{10}, you should know it well enough to say it as 282 million, 475 thousand, 249.

However, doing this same problem as a multiple of 10, it gets difficult to pronounce the proper places. What if you're given 700

^{10}? Instead of saying this answer, and possibly mis-naming the places, you should simply write out the answer of 28,247,524,900,000,000,000,000,000,000. You still get credit for doing this problem in your head, and you don't have the worry of mis-naming the place values!

Next, I'm going to teach you how to determine when two exponential expressions will give the same answer. Why is that important? Because it can help you turn seemingly difficult problems into ones you already know!

## Changing One Exponential Equation Into Another

Pop quiz time! What is 8^{2}? How about 4

^{3}? And finally, what's the answer to 2

^{6}?

You've probably realized that all 3 answers are the same. Is this just a coincidence, or is there a pattern here? You'll be glad to know, it's the latter.

Let's start with 2

^{6}. When we say 2

^{6}= 64, we're using a shorthand way of saying this:

2 * 2 * 2 * 2 * 2 * 2 = 64

Let's see what happens when we isolate pairs of 2s with parentheses (In mathematics, operations inside parentheses are performed before operations outside the parentheses), and then simplify them:

(2 * 2) * (2 * 2) * (2 * 2) = 64

Each of the (2 * 2) pairs can also be stated as 2

^{2}, so we'll change that:

(2

^{2}) * (2^{2}) * (2^{2}) = 64Interestingly, we still have a repetitive equation, so it can also be stated with exponents as:

(2

^{2})^{3}= 64Since the parentheses above denote that first the (2

^{2}) should be solved, we'll perform that before taking the result to the 3rd power. 2

^{2}is 4, so we can finish by writing it like this:

4

^{3}= 64Now you see why 2

^{6}and 4

^{3}give the same answer. Let's go through 2

^{6}again, but this time putting parentheses around three 2s, instead of just two:

(2 * 2 * 2) * (2 * 2 * 2) = 64

(2

^{3}) * (2^{3}) = 64(2

^{3})^{2}= 648

^{2}= 64This should help you see why 8

^{2}, 4

^{3}, and 2

^{6}all give the same answer. It also gives a glimpse of an interesting, and useful, pattern.

What pattern? Take a look at these three equations we've already determined are equal:

2

^{6}= (2^{2})^{3}= (2^{3})^{2}Since 6 = 2 * 3, isn't it interesting that taking a number to the sixth power can be broken up into two operations involving a 2nd and 3rd power? Let's see if this pattern holds up.

What is 2

^{10}? Yes, it's 1,024. We'll break the exponent of 10 into 2 * 5. So, if the pattern holds, then 2

^{10}should be the same as (2

^{2})

^{5}. This is simplified as 4

^{5}, and if you recall properly, 4

^{5}is indeed 1,024!

In other words, if you have an even exponent, you can divide the exponent by 2, take the base number to the 2nd power, and still get the same answer. For example, 3

^{4}is easily converted into (3

^{2})

^{2}, or 9

^{2}.

This doesn't just work for even powers, either. If your exponent is a multiple of 3, just divide the exponent by 3, and take the base number to the 3rd power. 2

^{9}is the same as (2

^{3})

^{3}, or 8

^{3}.

Hopefully, you get the general idea by now. An exponent that is a multiple of 4, can be divided by 4, as long as you take the base number to the 4th power to compensate. This holds true for exponents that are multiples of 5, 6, 7 and every other number out there. The mathematical way of saying this would be that X

^{(Y * Z)}= (X

^{Y})

^{Z}.

## Putting This to Use

Make sure you understand how to convert one exponential equation into another equivalent one, because I'm going to show you how to use this to go beyond the 10th power next!## Powers Beyond the 10th

I'm going to take you out of your comfort zone now. What is 2^{14}? Your first response is probably that you have no idea, as you only know the powers of 2 up to the 10th. Would you believe that you already know the answer?

Using the idea from the previous page, let's change it into another equation that we do know. 14 = 2 * 7, so how about (2

^{2})

^{7}? That's 4

^{7}, which you already know is 16,384. Congratulations, you just figured out 2

^{14}in your head!

Using this approach, you can now take 2 to all the even powers up through 20, since 2

^{20}= 4

^{10}. The shortcut here is to divide the exponent by 2, and change the base to 4. 2

^{16}? 4

^{8}! 2

^{12}? 4

^{6}! Do you see how quickly you can change difficult problems into easy ones?

## Taking 2 to Powers Which are Multiples of 3

Since 2^{3}= 8, and you know your powers of 8, you can also handle exponents that are multiples of 3 as far as 2

^{30}! Can you turn 2

^{21}into an equation that's easier to deal with? If you said 8

^{7}, you're right.

At this point, you should clearly understand how this conversion is being done, but just to reinforce the process, all you have to do is realize that 21 is 3 * 7, so 2

^{21}= 2

^{(3 * 7)}= (2

^{3})

^{7}= 8

^{7}. That is the mathematical explanation, however, not the thought process. Your thought process should be 21 is 3 times 7, so I'll change the 2 to an 8, and take it to the 7th power. Hey! 8

^{7}is 2,097,152!

Remember that story I told you about Dr. Solomon Golomb doing 2

^{24}in his head? This is the same method he used to work that out. He didn't know 2

^{24}, but he knew enough to convert it to 8

^{8}, which he knows (and you now know) is 16,777,216.

This does bring up a minor point. If you're given the problem 2

^{12}, you could convert it either 4

^{6}or 8

^{4}, and still get the right answer. Although it's ultimately personal preference, when faced with 2 to a power that's a multiple of 6, I always divide the exponent by 3, and change the 2 to an 8. Yes, 2

^{18}is 4

^{9}, but I'll think of it first as 8

^{6}.

## Taking 3 to Even Powers

Thanks to knowing your powers of 9, and the fact that 9 is 3^{2}, you can take 3 to any even power up to 20 using this same process. You're probably already ahead of me, but I'll explain anyway.

Let's take 3

^{12}. That's (3

^{2})

^{6}, or 9

^{6}(531,441, by the way). What's 3

^{20}? You should be able to convert it to 9

^{10}, and get the answer with little trouble.

## Putting This to Use

In subjects like computers and probability, powers of 2 and 3 come up remarkably often, so an ability to do these powers in your head can come be both handy and impressive. Practice these at the exponent quiz page, by using the Powers Up to 30 button.What could possibly be next? Instead of taking smaller numbers up to higher powers, we're going to reverse the process, and take larger numbers to smaller powers.

## Bases Larger than 10

A few pages back, I told you to make sure that you know your squares and cubes cold. That wasn't just to make sure you were able to recall them. It was also for this additional feat.Since you're now confident with problems up to 2

^{30}and 3

^{20}, here's a new kind of problem for you: What is 49

^{5}?

Before you give up, ask yourself whether 49 sounds familiar. It should, as it's the answer you memorized to 7

^{2}. If I rephrase the 49

^{5}problem as (7

^{2})

^{5}, that should become a more familiar problem. You should be familiar enough with this process now, that changing it into 7

^{10}, and then 282,475,249, should be no problem.

If you recognize the base as the square of a number from 1-10, and you know the square root (If A

^{2}=B, A is the square root of B) you can use this idea to take square numbers up to their 5th power. Let's try 81

^{4}. Which number, when squared, totals 81? Hopefully, you were able to recall that 81 equals 9

^{2}. From there, turning the problem into (9

^{2})

^{4}, or 9

^{8}, should pose few problems.

You can only take squares up to their 5th power, because any problem of the form (X

^{2})

^{5}is X

^{10}. However, being able to handle problems like 81

^{5}in your head is no small feat.

## Cubical Bases

We can also apply the same idea to cubes. Since (X^{3})

^{3}works out to X

^{9}, you won't be able to take cubes any higher than their 3rd power.

First, make sure you know all the cubes, and their respective cube roots (If A

^{3}=B, A is the cube root of B). 512? That's 8

^{3}. 343? That's 7

^{3}. 729? That's 9

^{3}. Don't forget the smaller ones, too. Sure, 64 is 8

^{2}, but it's also 4

^{3}.

If you feel ready, how about 343

^{3}? That's 7

^{9}, as you should know by now. How about the seemingly scary 729

^{2}? Yep, it's just a disguise for 9

^{6}. Don't stop there, however! You shouldn't forget to give the answer, which is 531,441.

## 4th Power and 5th Power Bases

Optionally, if you've become comfortable enough with the numbers to recognize 4th powers and 5th powers, to the point where you can quickly recall their respective 4th and 5th roots (As you probably already know, if A^{4}=B, A is the 4th root of B, and if A

^{5}=B, A is the 5th root of B), then you can handle squaring those 4th and 5th powers! Since (X

^{4})

^{2}is X

^{8}, and (X

^{5})

^{2}is X

^{10}, that's as far as you'll be able to with them.

Let's try a warm up with 1,296

^{2}(That's just a warm up?!?). Do you recognize that as a square (well, it is 36

^{2}, but that's not much help here), a cube, a 4th power, or a 5th power? You should recall it as 6

^{4}, so 1,296

^{2}becomes 6

^{8}(this should be 2nd nature by now), which you know to be 1,679,616!

The toughest one you can face is 59,049

^{2}. What root can you use? To what does the problem simplify? If you can change this into 9

^{10}in your head, and get the answer that way, you're probably ready to handle anything.

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