Negative Exponents

[jakep069]

I have just started a bridging / enabling program in math. I am an older bloke and feel like a dummy for asking, probably, such a basic question. I am starting a degree and am doing math just for fun because I like it. I can do harder things, but I had to teach myself, so I don't understand the basic principals and theory and have trouble understanding - WHY ! Because the sad thing is I have easier questions I need clarification on, on the basic theory, but am embaressed to ask.

Blue stuff represents defenitions:

(a^3.t / b^2.x)^-2 : (a/b)^-n

? (a^3.t)^-2 / (b^2.x)^-2 : a^-n / b^-n


? {1 / (a^3.t)^-2} / {1 / (b^2.x)^2} : a^-n = (1/a^n) / (1/a^n)

? {1 / (a^3.t)^-2} / {(b^2.x)^2 / 1} : a / b ÷ a / c = a / b x c / a = c / b

? (b^2.x)^2 / (a^3.t)^2 : from above = c/b

= b^4.x^2 / a^6.t^2 : (a^m)^n = a^mn
How's that ?

Thanks

Jason

 

[Deleted member 4993]

I have just started a bridging / enabling program in math. I am an older bloke and feel like a dummy for asking, probably, such a basic question. I am starting a degree and am doing math just for fun because I like it. I can do harder things, but I had to teach myself, so I don't understand the basic principals and theory and have trouble understanding - WHY ! Because the sad thing is I have easier questions I need clarification on, on the basic theory, but am embaressed to ask.

Blue stuff represents defenitions:

(a^3.t / b^2.x)^-2 : (a/b)^-n

? (a^3.t)^-2 / (b^2.x)^-2 : a^-n / b^-n


? {1 / (a^3.t)^-2} / {1 / (b^2.x)^2} : a^-n = (1/a^n) / (1/a^n)

? {1 / (a^3.t)^-2} / {(b^2.x)^2 / 1} : a / b ÷ a / c = a / b x c / a = c / b

? (b^2.x)^2 / (a^3.t)^2 : from above = c/b

= b^4.x^2 / a^6.t^2 : (a^m)^n = a^mn
How's that ?

Thanks

Jason

For quick review of laws of exponents - with worked examples - go to:

http://www.purplemath.com/modules/exponent.htm

 

[Denis]

You got from:

> (a^3.t / b^2.x)^-2

to:

> b^4.x^2 / a^6.t^2

KEERECT! Good work, Jason :idea:

Here's a "kwickie way":
First, your expression needs another set of brackets: (a^3.t / (b^2.x))^-2 ; agree?

RULE: a^(-p) = 1 / a^p

so (a^3.t / (b^2.x))^-2 = 1 / (a^3.t / (b^2.x))^2

square that: 1 / (a^6.t^2 / (b^4.x^2))

complete: b^4.x^2 / (a^6.t^2) ; Roger!

By the way, don't expect too much sympathy from me:
I'm 67 and got back to maths some 7 years ago, after being away some 40 years :wink:

 

[Deleted member 4993]

I'm 67 :wink:

... and still chasing loose pucks on ice... :lol: :lol: :lol:

 

[Denis]

I'm 67 :wink:

... and still chasing loose pucks on ice... :lol: :lol: :lol:

Well, better than dressing up in white surgeon's uniforms and trying
to prevent a little bowled ball from hitting innocent little wickets :twisted:

 

[jakep069]

Thank you all very much.

However, this is where my lack of basic knowledge comes in. I went to the, great, http://www.purplemath.com/modules/exponent.htm - site and have a general question.

The question on the site askes:

Write x^–4 using only positive exponents.

That's easy. Anything raised to a -pwr is changed to a +pwr, but over 1. That's all good and well because I 'just do it'. However, I don't understand the next part in that explanation that states:

1X^-4 / 1 : Where did the 1s come from and why?

Is it just because is is showing that any number x is just 1(x), and that any number x is just x/1 being x?


Also, Just thought I would add some more and edit this post before anyone replied. As per my explanation for the above for a problem such as:

2x^-2 My understanding is that it should now read (for my ease) as (2)(x)^-2 such that when applying the 1/a^n rule from above the 2 takes the place of the 1 in the (above) 1x^-4 in the numerator and the x^-2 moves to the denominator as the 'base'

 

[Denis]

> That's easy. Anything raised to a -pwr is changed to a +pwr, but over 1. That's all good and well because I 'just do it'.

PROVE it to yourself: use your calculator to get say 5^(-3); then try 1 / 5^3

> However, I don't understand the next part in that explanation that states:
> 1X^-4 / 1 : Where did the 1s come from and why?
> Is it just because is is showing that any number x is just 1(x), and that any number x is just x/1 being x?

Basically, YES!
Often, including the "1" makes it easier for a teacher to "demonstrate" something.

 

[Denis]

Also, Just thought I would add some more and edit this post before anyone replied. As per my explanation for the above for a problem such as:

2x^-2 My understanding is that it should now read (for my ease) as (2)(x)^-2 such that when applying the 1/a^n rule from above the 2 takes the place of the 1 in the (above) 1x^-4 in the numerator and the x^-2 moves to the denominator as the 'base'

Yes.
2x^-2 is really 2^1 times x^-2

IF it was 2^-1(x^2), then x^2 / 2 (or x^2 / 2^1): mix you up well?!

 

[jakep069]

G'day

I have another question to put to you, but I have been noticing that some questions asked in the difficult X^3 / a^2 form have been answered in the actual form as you would write it by hand. Is there a majic program that you all have that I can use to post in a more visably pleasing format?

Thanks

Jason