laws of exponents

[Luke55]

Hi i've just started algebra and now on the laws of exponents, i thought i understood but guess not. Which law is this and why does both convert into fraction form and then the number 1?

(x)(x^-1)

(x²)(x^-2)

Thanks

 

[Luke55]

sorry i thought it will show the exponents, basically the first one is x with negative 1 exponent and the second is x with a negative 2 exponent. Hope you understand...

 

[MarkFL]

Consider the following rule for exponents:

[MATH]a^b\cdot a^c=a^{b+c}[/MATH] where \(a\ne0\).

And so using this rule, what do you get for your two expressions?

 

[Luke55]

I think the first would be a and add the two exponents together and then you can multiply. Not sure about the a = 0

 

[MarkFL]

As long as both exponents are positive, then we can have \(a=0\), but \(0^b\) is only defined for \(0<b\).

So, let's look at the first one:

[MATH]x\cdot x^{-1}=x^1\cdot x^{-1}=x^{1-1}=?[/MATH]

 

[Luke55]

Im really sorry but im lost

 

[Steven G]

Let's look at another example first so that we can see the rule.

a³*a²
By definition this means a³*a²=(a*a*a)(a*a) = a*a*a*a*a = a⁵. Now how to we get the power 5. Basically we added the 2 and 3. So let's do that with your problem.

Note that x=x¹
For the first problem what is (1) + (-1)? Whatever that is, it will be the power of a. Can you finish up?

 

[HallsofIvy]

The rule that "aⁿa^m= a^n+m" follows from "counting 'a's" as Jomo showed.

Of course that method cannot be used if n or m are not positive integers since we cannot have 0 'a's or -1 'a's.

But then we are free to define other powers of a any way we want. But it would be nice to define those other powers so that nice law, aⁿa^m= a^n+m, is still true for other numbers.

0 has the property that, for any n, n+ 0= n. So we want to have aⁿa⁰= a^{n+ 0}= aⁿ. As long as a is not 0, we can divide both sides by aⁿ. We can have that same rule true as long as a⁰= 1.

For any positive n, there exist -n such that n+ (-n)= 0. aⁿa^-1= a^{n- n}= a⁰= 1. Again, dividing both sides by aⁿ (again requiring that a is not 0) we have a^-n= 1/aⁿ.

So x(x^-1)= x^1-1= x⁰= 1.

And x²(x^-2)= x^{2- 2}= x⁰= 1.

 

[Luke55]

Thank you i now understand, it was quite easy when i think about it now...