If you rewrite $(3^{12})(3^{-12})$ in the form $3^n$ what does it equal? What is the intuition behind it?

Do exponents cancel each other out so it is just $3$? or do the negatives cancel out $((-x) \cdot(-x)=x)$ so it is $3^{24}$? Or is it something else?

  • $\begingroup$ The exponents add, so $+12-12 = 0$ and the result is $3^0$. $\endgroup$ – Null Oct 21 '14 at 5:43
  • $\begingroup$ Thank you Why does 3 to the power of 0 equal 1 btw? $\endgroup$ – Ray Kay Oct 21 '14 at 5:50
  • $\begingroup$ Here is a good explanation. $\endgroup$ – Null Oct 21 '14 at 5:52
  • $\begingroup$ Do you know, Ray, that $3^{-12}$ is the same thing as $1/(3^{12})$? Does that help you see what's going on? $\endgroup$ – Gerry Myerson Oct 21 '14 at 6:26
  • $\begingroup$ see mathontrack.comze.com/exponentials2.html for a good explanation of several exponent rules (including why $a^0=1$) $\endgroup$ – John Joy Oct 21 '14 at 15:31

Consider what negative exponents represent. When we write $\ x^{ - a} $ we mean the same thing as $\ \dfrac{1}{x^a}$

Therefore, the multiplication of $\ x^{a} * x^{ - a} $ is equal to the multiplication of $\ x^{a} * \dfrac{1}{x^a} $

Which gives us $\ \dfrac{x^a}{x^a} $

We know this to be 1, as the division of a number by itself yields one.

A way to visualize this is to consider what the expanded forms of each power $\ x^{a} $ would be $\ x$ times itself a times. That is, x * x * x...* x * x and so on.

Consider if a equals 1. Then we have $\ x^{1} $, which evaluates to x.

If a equals 0, then the exponent evaluates to 1. The proof for this lies in division of exponents. If we have two exponents with the same base, let's call this base h, then $\ \dfrac{h^d}{h^e} $ (where e and d are two powers) is equal to $\ h^{d-e} $.

This is because $\ \dfrac{h^d}{h^e} $ can be written as

$\ \dfrac{h*h*h...h*h*h}{h*h*h...*h*h } $

where there are d h's multiplied on the numerator and e h's multiplied on the denominator

So, when we cancel out the h's, we divide out e h's from the d h's on top, leaving us with (d-e) h's multiplied together.

Consider what happens when d=e.

Well, then $\ \dfrac{x^d}{x^e} $, which is equal to $\ x^{d-e} $ evaluates to $\ x^0 $.

Note that $\ \dfrac{x^d}{x^e} $ is known to equal 1, because we know that a number over itself is 1. (Remember: d=e)

Therefore we have established that $\ x^0 $ is equal to 1.

Now consider what happens when our exponent, a is less than zero.

To solve this, consider what happens when we "take away" 1 from a positive a value. Well, we are simply taking one x away (that is, dividing it out).

For instance, consider the case of $\ x^5 $. This can be expressed as $\ x*x*x*x*x $. If we "take away" an x from this multiplication (i.e., divide an x), we are left with a product of four x's. In other words, we are left with $\ x^4 $.

Now extend that definition for $\ x^0 $. By our logic, to arrive at $\ x^{-1} $ we simply divide out another x from $\ x^0 $. We have found that $\ x^0 $ is equal to 1. Accordingly, 1 divided by x is our value for $\ x^{-1} $.

Keep going, to find, say, $\ x^{-5} $, we are just dividing four more x's from $\ x^{-1} $. (Remember: every time we decrease the power by one, we are dividing out an x)

We could extend a similar logic to fractional exponents, but I feel that may be extraneous in the light of the question.

EDIT: Similar to how you can divide two exponents with the same bases by subtracting the exponents, you can multiply to exponents with the same bases by adding them. Likewise, we can reach the answer to your question directly through that method (this is discussed in the other post, as well). Remember though, where this comes from. We are really multiplying ((3)(3)(3)(3)...(3)(3)(3)(3)) * ((1/3)(1/3)(1/3)(1/3)...(1/3)(1/3)(1/3)(1/3)). Because 1) both fractions have the same base and 2) the negative and positive powers have equal magnitudes (|12|=|-12|). Because there are as many negative exponents there are positive, there are as many 3's as there are (1/3)'s. So, each of the 1/3 's cancels out with each of the 3's, leaving only 1's. And the product of 1's is always 1.


we have $3^{x}\cdot 3^{-x}=3^{x-x}=3^0=1$ or $\frac{3^x}{3^{x}}=1$


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