Laws of Exponents

Published: Dec 25th, 2020

Learning Objectives
By the end of this section, you should be able to:
1. Simplify and evaulate expressions containing exponents using the laws of exponents. 

Well, it's time you get acquainted with the laws around here...

... the laws of exponents, that is!

Unfortunately for you, these rules have to be committed to memory sad. However, as I've said before, I believe that math should be about intuition rather than memorization, so let's try to put a few of these ideas into words:

1. \(a^m \times a^n = a^{m+n}\)

This one might be hard to conceptualize, but let me give you an example:

\(2^3 \times 2^5 = ?\)

Another way we can write these two terms is as \(2 \times 2 \times 2\) and \(2 \times 2 \times 2 \times 2 \times 2\), respectively. Now, if we multiply these two terms, how many times are we multiplying by 2? Eight times? Interesting, because 8 is also 3+5 (the sum of the exponents!). In other words, when two exponents have the same base and are multiplied, the exponent of the product is the sum of the exponents.

2. \(a^m \div a^n = a^{m-n}\)

Now, let's keep our original expression, but change the symbol: \(2^3 \div 2^5 = ?\)

Write this as a fraction: \(\frac{2\times2\times2}{2\times2\times2\times2\times2}\)

Common factoring out the 2s from the numerator and denominator, what are we left with?

\(\frac{1}{2\times2}\)

Another way we write this is:

\(\frac{1}{2^2} = 2^{-2}\)

Note that 3-5 = -2 (subtracting the exponent of the denominator from the numerator)! 

Therefore, we can skip these steps simply but subtracting the exponent of the denominator from the numerator, when the bases are the same in both terms.

3. \((a^m)^n\) = \(a^{mn}\)

Put this in number form: 

\((5^2)^2\), for example.

Rewrite this as a product: 

\((5\times5)\times(5\times5)\)

How many times are we multiplying by 5? 4 times!

Note that \(2 \times 2 = 4\), as well. 

Therefore, the exponent of an exponent is the product of the exponents.

Further, I also want to discuss zero and negative exponents. Anything, except for 0, when raised to the power of 0, is equal to 1.

\(a^0 = 1\), if \(a \neq 0 \)

What about negative exponents? Those sound hard, right? Well, if you've been following along, in Lesson 7: Four Basic Functions, we stated that the inverse function is \(f^\color{red}{-1}(x)\). For the function, \(f(x)=x\), the inverse would be \(x^{-1}\), or \(\frac{1}{x}\). Little did you know, but this is a negative exponent! Now, what if the negative value of the exponent is greater than 1?

Solve \(\frac{a^3}{a^5}\)

Looking at our second property, we know that we should subtract these exponents: 

\(= a^{3-5}\)

\(= a^{-2}\)

What if we don't want a negative exponent? Well, we express it as an inverse and remove the negative!

\(\frac{1}{a^2}\)

What about an INCEPTION case? The inverse of an inverse?

\(\frac{1}{x^{-1}}\)

Here comes the inception part:

\(\frac{1}{\frac{1}{x}}\)

Whoa! Let's try to simplify that by carrying out the first division (we discuss the division of rational expressions in Lesson 12)

\(=1 \times \frac{x}{1}\)

\(=x\)

With that out of the way, let's go through some examples that may help you develop a sense for these rules and minimize your memorization troubles. 

Solve \(5^2 \times 5^3 \times 5^4 \times 5^5\)

Whoa! I never learned up to there in multiplication tables.

Applying one of our laws, however, this turns into an addition problem:

\(= 5^{2+3+4+5}\)

\(= 5^{14}\)

That wasn't so bad. No need to evaluate for these questions; just get it to a simplified exponent. 

Solve \(\frac{9^2\times9^5\times9^{13}}{9^4}\)

\(=\frac{9^{2+5+13}}{9^4}\)

\(=9^{16}\)

Solve \((\frac{3^2\times2^5}{2^3\times3^4})^3\)

\(=(3^{2\color{red}{-4}}\times2^{5\color{red}{-3}})^3\)

\(=(3^{-2}\times2^2)^3\)

\(=3^{-6}\times2^6\)

Solve \(\frac{(2ab)^3c}{(4a^2c)^2}\)

\(=\frac{8a^3b^3c}{16a^4c^2}\)

\( =\frac{b^3}{2ac}\)

Solve \(\frac{(a^x)^4(a^{2x+3y})^3}{(a^{y-2x})^2}\)

\(=\frac{(a^{4x})(a^{3(2x+3y)})}{a^{2y-4x}}\)

\(=\frac{(a^{4x+6x+9y})}{a^{2y-4x}}\)

\(=\frac{a^{10x+9y}}{a^{2y-4x}}\)

\(=a^{10x\color{red}{+4x}+9y\color{red}{-2y}}\)

\(=a^{14x+7y}\)

This is another application section; not much to teach, but quite a lot to practice. Be sure to have a stab at the practice problems!

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