Tag Archives: elementary math

Exponent Misery

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Kory Graham of @korytellers fame, sent me this tweet almost a month ago

and as usual, I finally am getting around to it.

Now, there are typical ways to start to understand the “rules of exponents” and this can be found in almost any Algebra or pre-Algebra textbook. Assuming that these rules were already explained and are available, I will try a different approach to explain what is going on. I’ll go through the problems with the intent to help that the rules be understandable. Two things for the student to note and keep in mind when reading this:

  1. Read this again. It’s ok if it doesn’t make sense the first time. The first reading through of mathematics is so that the next time(s) you read this, you know what to expect. The first time around, you’re still adjusting to all the surprises.
  2. All the problems are the same! Try to see how they are. 🙂

Ok, in all of these problems we are asked to simplify our answers and only leave positive exponents. Let’s just start and we’ll seeing what simplifying means.

Problem 1

$$-3xy^{3}\cdot 2xy^{3}$$
One of the things that can be confusing is what the notation means. Typically to read this out loud, we are taught to say it as “negative 3 times \(x\) times \(y\) cubed times 2 times \(x\) times \(y\) cubed”. This is very much a left-to-right way of reading the expression. But math symbolism doesn’t have to be read left to right.

Now, a tricky part is, what does “\(x\) times \(y\) cubed” mean? What’s the difference between $$(xy)^{3}$$ and $$xy^{3}$$? Can you see the difference in how it is written? (Notice the parentheses.) But if we were to read these statements out loud, we should read them like this: \((xy)^{3}\) is read as “the cube of \(x\) times \(y\)” or “the cube of the product of \(x\) and \(y\)”. Whereas \(xy^{3}\) may be better read as “\(x\) times the cube of \(y\)” or “the product of \(x\) and the cube of \(y\)”.

If that doesn’t make any sense from a language standpoint, try language more familiar: What’s the difference between “the money of Joe and Jane” vs “Joe and the money of Jane”? If you can understand the difference in these sentences, then you’re just a hop, skip, and a jump away from understanding the difference between “the cube of \(x\) times \(y\)” and “\(x\) times the cube of \(y\)”. (Compare and contrast the structure of the sentences!)

Ok, enough of that, and let’s continue to the problem. Simplify
$$-3xy^{3}\cdot 2xy^{3}$$

Let’s start with the answer:
$$-3xy^{3}\cdot 2xy^{3} = -6x^{2}y^{6}$$
This is the answer that was written, but it’s not clear from the work how we arrived to it. So, here’s a step-by-step breakdown that may help in case we just got lucky. Note that there are several ways to arrive at the final simplification.

First, \(-3xy^{3}\) can be written as \(-3xyyy\) (because only the \(y\) is cubed!). Next, \(2xy^{3}\) can be written as \(2xyyy\). So, if we were going to multiply these two expression together, we would have \(-3xy^{3}\cdot 2xy^{3} = -3xyyy\cdot 2xyyy = -6xxyyyyyy = -6x^{2}y^{6}\).

Alternatively, we can use our rules of exponents and recognize that when when the bases are the same, we can simply add exponents when multiplying. In other words, \(x^{m}\cdot x^{n} = x^{m+n}\), or in our case \(-3xy^{3}\cdot 2xy^{3} = -3\cdot 2 \cdot x^{1}x^{1} \cdot y^{3}y^{3} = -6x^{2}y^{6}\)

If this makes sense, then Problem 2 will be a cinch.

Problem 2

$$m^{3}n^{2}\cdot 4m^{2}n^{2}$$
The work starts of correctly, except we divided! Oops! So, following the work that was written, we have \(m^{3}n^{2}\cdot 4m^{2}n^{2} = mmm\cdot nn \cdot 4 \cdot mm\cdot nn\) and rearranging however we likely because this is all multiplication, we have \(4 \cdot mmmmm \cdot nnnn = 4m^{5}n^{3}\). Or we could use the same exponent law from Problem 1 and see that this is \(m^{3}n^{2}\cdot 4m^{2}n^{2} = 4\cdot m^{3}m^{2}\cdot n^{2}n^{2} = 4m^{5}n^{4}\).

Now try these:

Like Problem 1 and Problem 2

Simplify \(4x^{3}r^{5}\cdot x^{6}r^{7}\) and simplify \(p^{4}y^{9}\cdot 3p^{2.2}y\) (yes that’s a decimal in the exponent! Does that change anything? How? and why?)

Problems 3 and 4

Problems 3 and 4 are similar to Problems 1 and 2 and the boxed answers given are pretty much correct, they just have to be taken one step further for “simplification”. One way to understand “simplify” is “re-write the expression as compactly as possible where each term can stand on its own”. The only concern is in understanding the negative sign and how it is used.

So, if we have \(-m^{3}\) that can be written as \(-m\cdot m \cdot m\) (notice there is only one negative sign). If, however, we had \((-m)^{3}\) we could write this as \(-m\cdot -m \cdot -m\) (notice that there are three negative signs. If we wanted to read out loud \(-m^{3}\) we would read this as “the negative of the cube of \(m\)”. If we read \((-m)^{3}\) out loud, we would read this as “the cube of negative \(m\)”. Both of these cases, would still produce a final result of \(-m^{3}\), but how we get there would be different. (Remember that if we write \(-m\) that’s the same thing as writing \(-1\cdot m\), but because of convention (laziness), the one is omitted because “it is understood”.) (And, yes, for the nitpickers, I am aware of the hair-splitting we can do here.)

So, for Problem 3 we have to simplify $$-m^{3}n^{3}m^{3}n^{3}$$ and if we wrote this out in “long hand” we would have $$-mmm\cdot nnn \cdot mmm \cdot nnn$$ which would be simplified to $$-m^{6}n^{6}$$

Or we could use our exponent law and recognize that \(-m^{3}n^{3}m^{3}n^{3} = -m^{3}m^{3}n^{3}n^{3} = -m^{6}n^{6}\).

Problem 4 is no different. In fact, the answer written is correct, except the negative sign is just throwing us off. The boxed answer is $$-2m^{4}n^{7}3$$ which we would just simplify like so, \(-2m^{4}n^{7}3 = -2\cdot 3 \cdot m^{4}n^{7} = -6m^{4}n^{7}\).

Ok, now for Problem 5 through 8, starting with Problem 5.

Problem 5

Believe it or not, these problems are no practically no different from Problems 1 through 4! Why?

Let’s take a look at Problem 5. Simplify
$$(4m^{2})^{3}$$

This one can be a bit tricky because of the multiple exponents and the parentheses. But! If we defocus for a second we can look at this problem as “\((blah)^{3}\)”. And if “\(blah\)” is cubed, that just means \(blah \cdot blah \cdot blah\). But what is \(blah\)? Can you see that \(blah\) should be \(4m^{2}\)? And if we said that \(blah\) was cubed, then we have that \(4m^{2}\) should be cubed. So that means that \((4m^{2})^{3} = 4m^{2}\cdot 4m^{2}\cdot 4m^{2}\) and can you see how this is just like the Problems 1 through 4? Can you see that the final simplification is \(64m^{6}\)? (\(4\cdot 4\cdot 4 = 64\) and \(m^{2}m^{2}m^{2} = m^{6}\) by our exponent law.)

Now, try to use the same reason for Problems 6, 7, and 8. Here are the solutions to check against.

Simplify Problem 6
$$(2x^{4}y^{3})^{2} = 2^{2}x^{4\cdot 2}y^{3 \cdot 2} = 4x^{8}y^{6}$$

Simplify Problem 7
$$(2y)^{4} = 2^{4}y^{4} = 16y^{4}$$

Simplify Problem 8
$$(3mn^{3})^{3} = 3^{3}m^{3}n^{3\cdot 3} = 27m^{3}n^{9}$$

Notice that the exponent law here is $$(x^{n})^{m} = x^{n\cdot m}$$

Understanding this law can be made easier if we try to think of the term with in the parentheses as being raised to the \(m\) power. So, \((x^{2})^{3} = x^{2}\cdot x^{2}\cdot x^{2} = x^{6}\) because we are cubing \(x\) squared. And \((x^{3})^{2} = x^{3}\cdot x^{3} = x^{6}\) because we are squaring \(x\) cubed.

Warning! Warning! Even though the “pattern” may be that we “distribute the exponent” (not to be thought of that way!!), we have to recognize that this only holds when the terms inside the parentheses are all multiplied. Convince yourself that generally \((x + y)^{2} \neq x^{2} + y^{2}\). (Use \(x = 3\), \(y = 1\), for example.)

Problems 9 and 10

Both of these problems are done correctly! There are other ways of simplifying these, but the method used is correct. If we wanted to “strictly” stay within the lesson about exponents, we could do the following for Problem 9
$$\frac{4b^{3}}{2b^{2}} = 2^{2}b^{3}2^{-1}b^{-2} = 2^{1}b^{1} = 2b$$

A different question that could be asked to further test our understanding of negative exponents would be this one. Simplify
$$\frac{4b^{3}}{2b^{-2}}$$

See if you can figure out that the simplification should be \(2b^{5}\).

Happy Mathing! And remember, if it didn’t make sense the first time around, read it again! And read it slower! And as soon as you don’t understand something, ask yourself what was the last thing you did understand and go from there.