How do you simplify the following expressions?

[hexhall]

I'm only in 9th grade math, so please don't give me complicated calculus answers. And I prefer if you give steps on how to solve each step. Thanks in advance!

1. 5^-2/p
2. 3x^-2/y
3. (x^-5) (y^-7)
4. 8/2c^-3
5. (6a^-1) (c^-3)/d^0
6. (9^0) (y^7) (t^-11)

 

[PeterO]

I'm only in 9th grade math, so please don't give me complicated calculus answers. And I prefer if you give steps on how to solve each step. Thanks in advance!

1. 5^-2/p
2. 3x^-2/y
3. (x^-5) (y^-7)
4. 8/2c^-3
5. (6a^-1) (c^-3)/d^0
6. (9^0) (y^7) (t^-11)

Unusual that you post this on the Physics forum but ...

Simplifying in indice problems usually means either

collecting all terms in the numerator
or
expressing without negative indices,
and
perhaps evaluating numerical indices

I that order, number 1 becomes:

5^-2p^-1
or
1/(5²p)
or
1/(25p)

really depends which one of those you think is the most simplified.

Note: I typed in brackets in the 2nd and 3rd example lest you thought only the first part was in the denominator.
When writing by hand you can make a clear fraction with a large dividing line so it is clear both the 25 and p, for example, are in the denominator.

With indices, combining those indices with the same base usually constitutes simplification, but that does not apply to any of these examples.

 

[hexhall]

I still don't really understand the process in which you figured it out...

 

[PeterO]

I still don't really understand the process in which you figured it out...

Does that mean you don't understand negative indices?

That is a more basic problem if that is the case.

 

[hexhall]

Yes... Could you just explain it one more time, please?

 

[PeterO]

Yes... Could you just explain it one more time, please?

2^-3 = 1/2³

going out now won't have a further reply for 8 hours.

 

[hexhall]

2^-3 = 1/2³

going out now won't have a further reply for 8 hours.

Okay, well, that didn't really answer my question...? Is this a new problem? I don't think you get what I'm asking.

 

[emailanmol]

Hey,

See x^3 is like saying x*x*x.

On the other hand x^(-3) is like saying (1/x)*(1/x)*(1/x)

That is (1/x^3).

So 5^(-2)/p is like (1/5)*(1/5)*(1/p) so its (1/25p)

Is it clear??


In general x^a is like multiplying x with itself a times.

x^(-a) is like multiplying
(1/x) a times.

 

[emailanmol]

Try this site.Read as much as can :-)

http://www.mathsteacher.com.au/year10/ch08_indices/08_negative_indices/neg.htm


Read only the yellow table in this link. It will help a lot.

http://www.mathsisfun.com/algebra/exponent-laws.html

 

[PeterO]

Okay, well, that didn't really answer my question...? Is this a new problem? I don't think you get what I'm asking.

Not really sure even YOU know what you are asking.

In post#2 I gave you the 3 most likely answer to the first problem. Which one is "correct" depends on the context in which the questions were given.

As I said:

Some people think expressing indices with a denominator of 1 is it.

Some people think using only positive indices is it.

Some people think that evaluating every indice that has a number as a base is what is required.

What do you think is required?

 

[hexhall]

Hey,

See x^3 is like saying x*x*x.

On the other hand x^(-3) is like saying (1/x)*(1/x)*(1/x)

That is (1/x^3).

So 5^(-2)/p is like (1/5)*(1/5)*(1/p) so its (1/25p)

Is it clear??


In general x^a is like multiplying x with itself a times.

x^(-a) is like multiplying
(1/x) a times.

Hey, you answered one of my other questions! Thanks for explaining it into simpler terms. And thanks for the links too. I get it now