Algebra Chapter 1 Section 6: Properties of Exponents
In the first section of this lecture series, I touched on positive integers as exponents. Now I will delve further into the rules of exponents.
Let's say that we're multiplying two exponents, $x^{5}$ and $x^{2}$, together. Notice that $x^{2}=x \times x$ and $x^{5}=x \times x \times x \times x \times x$. This means that if we multiply those two values together, we have $x^{2} \times x^{5}=(x \times x) \times (x \times x \times x \times x \times x)=x \times x \times x \times x \times x \times x \times x$. This means that $x^{2} \times x^{5}=x^{7}$. This means that when you're multiplying two exponents together, so long as the two bases are the same, all that needs to be done is adding the two exponents together, $x^{2} \times x^{5}=x^{5+2}=x^{7}$.
We can use the same concept for division (quotients), $\frac{x^{5}}{x^{2}}$. Notice that this yields $\frac{x^{5}}{x^{2}}=\frac{x \times x \times x \times x \times x}{x \times x}$. Two of the x's from the numerator cancels with the two x's in the denominator, so this gives us $\frac{x^{5}}{x^{2}}=x \times x \times x=x^{3}$. Notice that the final answer happens to have an exponent which is equal to the difference between the two exponents in the initial expression, $3=5-2$. This is no coincidence; in fact, this can be generalized as The Exponent Rule of Quotients: $\frac{a^{m}}{a^{n}}=a^{m-n}$, so long as the bases are identical.
An obvious question is what would happen in The Exponent Rule of Quotients if n>m? Let's look in to that. If we have $\frac{x^{2}}{x^{5}}=\frac{x \times x}{x \times x \times x \times x \times x}$, then two x's from the denominator cancels with the two x's in the numerator, giving us $\frac{x^{2}}{x^{5}}=\frac{1}{x \times x \times x}=\frac{1}{x^{3}}$. If we wanted an alternative aesthetic of this fraction (having an implicit fraction instead of an explicit fraction), we could do it without loss of accuracy. The concept of $\frac{a^{m}}{a^{n}}=a^{m-n}$ still applies here so that $\frac{x^{2}}{x^{5}}=x^{2-5}=x^{-3}$. This means that $x^{-3}=\frac{1}{x^{3}}$.
A similar logic can be applied when m=n. If we have $\frac{x^{5}}{x^{5}}=\frac{x \times x \times x \times x \times x}{x \times x \times x \times x \times x}$, then all of the x's in the numerator and all of the x's in the numerator cancel out so that we have $\frac{x^{5}}{x^{5}}=\frac{1}{1}=1$. Using the subtraction principle, we find that $\frac{x^{5}}{x^{5}}=x^{5-5}=x^0$. Notice that this case has $x^{0}=1$. As it so happens, this applies to all finite values of x which is not 0.
Why not 0? The answer is simple: if you have $0^{n}$, then you will be multiplying a certain number of zeros together, and anything which is multiplied by zero is zero, so $0^n=0$ for all $n\ne0$. So when we have $0^{0}$, we have two contradictory statements; either it has to be 0 (from the base being 0) or it has to be 1 (from the exponent being 0). This means that we have $0=1$, which we know is false. So that means that the statement "$0^{0}$ has a solution" is false. That means that there is no solution to the expression $0^{0}$.
When we have an exponent on the outside of a parentheses, there are many different contexts, but I will cover just three of them here. When we have just a base with an exponent inside parentheses, and an exponent outside, such as $(x^{5})^{2}$, it is equivalent to $$(x \times x \times x \times x \times x)^2=(x \times x \times x \times x \times x) \times (x \times x \times x \times x \times x).$$Notice that there are now ten x's being multiplied together, so the result is $x^10$, which you'll notice is equivalent to $x^{5 \times 2}$. This can be generalized to $(a^{m})^{n}=a^{m \times n}$. This concept can be applied to fractions as well; when you have $(\frac{a^{m}}{b^{n}})^o$, it is equivalent to $\frac{a^{m \times o}}{b^{n \times o}}$, provided that $b\ne0$. This also applies to the fact that the exponent of the product of two numbers is equal to the product of two exponents, $(a \times b)^{n}=a^{n} \times b^{n}$.
We can use the same concept for division (quotients), $\frac{x^{5}}{x^{2}}$. Notice that this yields $\frac{x^{5}}{x^{2}}=\frac{x \times x \times x \times x \times x}{x \times x}$. Two of the x's from the numerator cancels with the two x's in the denominator, so this gives us $\frac{x^{5}}{x^{2}}=x \times x \times x=x^{3}$. Notice that the final answer happens to have an exponent which is equal to the difference between the two exponents in the initial expression, $3=5-2$. This is no coincidence; in fact, this can be generalized as The Exponent Rule of Quotients: $\frac{a^{m}}{a^{n}}=a^{m-n}$, so long as the bases are identical.
An obvious question is what would happen in The Exponent Rule of Quotients if n>m? Let's look in to that. If we have $\frac{x^{2}}{x^{5}}=\frac{x \times x}{x \times x \times x \times x \times x}$, then two x's from the denominator cancels with the two x's in the numerator, giving us $\frac{x^{2}}{x^{5}}=\frac{1}{x \times x \times x}=\frac{1}{x^{3}}$. If we wanted an alternative aesthetic of this fraction (having an implicit fraction instead of an explicit fraction), we could do it without loss of accuracy. The concept of $\frac{a^{m}}{a^{n}}=a^{m-n}$ still applies here so that $\frac{x^{2}}{x^{5}}=x^{2-5}=x^{-3}$. This means that $x^{-3}=\frac{1}{x^{3}}$.
A similar logic can be applied when m=n. If we have $\frac{x^{5}}{x^{5}}=\frac{x \times x \times x \times x \times x}{x \times x \times x \times x \times x}$, then all of the x's in the numerator and all of the x's in the numerator cancel out so that we have $\frac{x^{5}}{x^{5}}=\frac{1}{1}=1$. Using the subtraction principle, we find that $\frac{x^{5}}{x^{5}}=x^{5-5}=x^0$. Notice that this case has $x^{0}=1$. As it so happens, this applies to all finite values of x which is not 0.
Why not 0? The answer is simple: if you have $0^{n}$, then you will be multiplying a certain number of zeros together, and anything which is multiplied by zero is zero, so $0^n=0$ for all $n\ne0$. So when we have $0^{0}$, we have two contradictory statements; either it has to be 0 (from the base being 0) or it has to be 1 (from the exponent being 0). This means that we have $0=1$, which we know is false. So that means that the statement "$0^{0}$ has a solution" is false. That means that there is no solution to the expression $0^{0}$.
When we have an exponent on the outside of a parentheses, there are many different contexts, but I will cover just three of them here. When we have just a base with an exponent inside parentheses, and an exponent outside, such as $(x^{5})^{2}$, it is equivalent to $$(x \times x \times x \times x \times x)^2=(x \times x \times x \times x \times x) \times (x \times x \times x \times x \times x).$$Notice that there are now ten x's being multiplied together, so the result is $x^10$, which you'll notice is equivalent to $x^{5 \times 2}$. This can be generalized to $(a^{m})^{n}=a^{m \times n}$. This concept can be applied to fractions as well; when you have $(\frac{a^{m}}{b^{n}})^o$, it is equivalent to $\frac{a^{m \times o}}{b^{n \times o}}$, provided that $b\ne0$. This also applies to the fact that the exponent of the product of two numbers is equal to the product of two exponents, $(a \times b)^{n}=a^{n} \times b^{n}$.
That's the end of this section. If you have any questions, please leave them is the comments. Like and share this post if you found it helpful. And until next time, stay curious.
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To help get this lecture series come out with higher frequency, please donate to The Science of Life. This helps keep the information current and allows me to dedicate more time to this project instead of obtaining money through external means.
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