- #1

- 753

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Let 0 < a, an element of R, and x, y an element of R. Then:

(a^x)(a^y) = (a^(x+y))

a^(-x) = 1/(a^x)

Thanks.

- Thread starter Nusc
- Start date

- #1

- 753

- 2

Let 0 < a, an element of R, and x, y an element of R. Then:

(a^x)(a^y) = (a^(x+y))

a^(-x) = 1/(a^x)

Thanks.

- #2

- 753

- 2

A hint before midnight would be great!

- #3

- 753

- 2

Or maybe they are not so simple?

- #4

- 42

- 0

what class is this for? There is prolly many ways to prove this but some of the techniques might not be appropriate for all classes.

how about write

(aaaaa....a)(aaaaaaa......a) so we have a times itself x+y times

....x times.......... y times

=a^(x+y)

how about write

(aaaaa....a)(aaaaaaa......a) so we have a times itself x+y times

....x times.......... y times

=a^(x+y)

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- #5

- 5

- 0

Since 1(anything)=anything

so try multiplying by 1 where your 1= (a^x)/(a^x)

- #6

lurflurf

Homework Helper

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The second follows easily from the first by the case y=-xNusc said:

Let 0 < a, an element of R, and x, y an element of R. Then:

(a^x)(a^y) = (a^(x+y))

a^(-x) = 1/(a^x)

Thanks.

To handle the first you need to consider the definition you are using for a^x

for instance one possible definition is

a^x:=exp(x*log(a))

in which case the problem is reduced to showing

exp(x+y)=exp(x)*exp(y)

for this one must considerthe definition of exp(x)

one possible definition being

exp(x) is the unique function for which

exp(x) is a real number if x is a real numberjk

exp(x)*exp(y)=exp(x+y) if x and y are real numbers

limit x->0 [exp(x)-1]/x=1

This is a nice definition for this problem, of course if other definitions are used you will need to prove your statement other ways.

The other common definition of a^x is to let a^x be defined for rational numbers r then

a^x:=lim n->infinity a^r_n

where r_n is a rational sequence for which

lim n->infinity r_n=x

it is also possible to stay with

a^x:=exp(x*log(a))

and use other definitions for exp and log

- #7

- 753

- 2

(a^x)(a^y) = exp(x*ln(a))exp(y*ln(a)) = exp((x+y)*ln(a)) = exp(ln(a)^(x+y)) = a ^ (x+y)

Okay there. Now the second one?

exp(-x*ln(a)) = exp(ln(a)^-x) = a ^-x = 1/a^x

I'm not too sure about the second one...

Okay there. Now the second one?

exp(-x*ln(a)) = exp(ln(a)^-x) = a ^-x = 1/a^x

I'm not too sure about the second one...

Last edited:

- #8

lurflurf

Homework Helper

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The second is a special case of the firstNusc said:(a^x)(a^y) = exp(x*ln(a))exp(y*ln(a)) = exp((x+y)*ln(a)) = exp(ln(a)^(x+y)) = a ^ (x+y)

Okay there. Now the second one?

exp(-x*ln(a)) = exp(ln(a)^-x) = a ^-x = 1/a^x

I'm not too sure about the second one...

a^x*a^-x=a^(x+(-x))=a^(x-x)=a^0=1

hence

a^-x-1/a^x

just remember what is needed to rigorize this

1)take as a definition

a^x:=exp(x*log(a))

2)show that definition is equivelent other definitions if needed

3)define exp and log as desired

4)show these definitions are equivelent other definitions if needed

5)show that exp(x+y)=exp(x)*exp(y)

6)show that 5 and 1 together can be used to prove the desired result

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