Question

prove that, for all nonzero a, and b

(a * b)^x = a^x * b^xprove that, for all nonzero a, and b

(a * b)^x = a^x * b^x@Mathematics

Answer 1

(ab)^x = e^(xln(ab)) =e^(xln(a)+xlnb) =e^(xln(a))*e^(xln(b)) = a^x * b^x

Answer 2

(ab)^x = (ab)(ab)...(ab) , rearrange the a's, and b's

Answer 3

that proof works.

Answer 4

(ab)^x = (ab)(ab)(ab)...= (aaa...)(bbb...)

Answer 5

but this only works for positive integer x, so xavier's proof is better

Answer 6

but using his proof would mean that I have to know what ln means.

Answer 7

ln is the inverse of e^x

Answer 8

Yeah. Some properties of logs are used and the fact that e^(a+b)=e^a*e^b

Answer 9

ok, well then , how can you show that (ab)^(1/3) = (a)^1/3 * b^(1/3)

Answer 10

xavier, youre proof also works for irrational numbers like pi

Answer 11

:):):):):):)

Answer 12

i love pi

Answer 13

It assumes a lot though. Pecan pi :)

Answer 14

it assumes things like least upper bound property , but thats fine because real numbers have completeness property

Answer 15

agdg, are you doing this for a class?

Answer 16

yes

Answer 17

we define number as the equivalence class of sets which are 1-1 correspondance with it

Answer 18

a=-1,b=-1,x=1/2

Answer 19

\[(ab)^x = ((ab)^1)^x = (a^1b^1)^x = a^x * b^x\]meh my proof is lame... but I'm getting there :-(

Answer 20

that doesnt follow

Answer 21

it is hard to prove something that isn't true...you need to put more restrictions on a and b. see my example

Answer 22

right

Answer 23

In step 2 to 3 of your proof you used what you are trying to prove

Answer 24

(ab)^x = e^(xln(ab)) =e^(xln(a)+xlnb) =e^(xln(a))*e^(xln(b)) = a^x * b^x,
so a>0 , b>0 here, otherwise it does not work

Answer 25

because ln a is not define for a <=0

Answer 26

I can already prove that \[(ab)^{-1} = a^{-1}b^{-1}\]

Answer 27

yeah thats easy

Answer 28

i have an interesting question
what is
-8^(2/3)

Answer 29

it is -4

Answer 30

4?

Answer 31

-4

Answer 32

order of operations i think

Answer 33

I learned that exponentiation takes precedence over unary minus the hard way :(

Answer 34

heres another zinger

Answer 35

(-8)^(4/2)

Answer 36

64

Answer 37

:)

Answer 38

if we do it as
[(-8)^(1/2)]^4 that is not defined

Answer 39

but ((-8)^4)^(1/2) is defined

Answer 40

if you allow complex numbers it comes out the same though.

Answer 41

Do parenthesis first?

Answer 42

oh nice

Answer 43

:( won't anyone prove for me that (ab)^x = a^x * b^x ? :(

Answer 44

\[[(-8)^{\frac{1}{2}}]^4=[i\sqrt{8}]^4=i^4\cdot 64=64\]

Answer 45

I already did

Answer 46

but you said it won't work for x in (-1, 0) U (0, 1)

Answer 47

(ab)^x = (ab)(ab)...(ab) x times
rearrange by commutativity of numbers
(aa....a)(bb...b) where each of them are x times
a^x*b^x

Answer 48

lets just change my question to include for all positive reals x, use Xavier's proof, and call it a day?

Answer 49

actually it does, for instance
(ab)^.3 = (ab)^(3/10) = [(ab)^3]^(1/10)

Answer 50

so it works for decimals

Answer 51

it just doesnt work for irrationals. no proof will unless you use least upperbound property

Answer 52

but how do we find (a^3 * b^3)^0.1 ?

Answer 53

and show that it is equal to a^0.3 * b^0.3

Answer 54

(a^3 * b^3)^0.1 = (a^3 * b^3)^(1/10) = [(a^3 * b^3)^1]^(1/10)

Answer 55

lol

Answer 56

:):):):)

Answer 57

I will borrow my professor's answer book and see the one-liner answer for myself :-D

Answer 58

so we have to find fractions in general
(ab)^(1/c) = a^(1/c)*b^(1/c) , prove that

Answer 59

but thats easy , 1/c = c^-1

Answer 60

ok i got it

Answer 61

yay

Answer 62

(a^3 * b^3)^0.1 = (a^3*b^3) ^(10^-1) = [(a^3 * b^3)^10]^-1

Answer 63

mistake there.

Answer 64

(a^3 * b^3)^0.1
(a^3*b^3)^1/10
= (a^3*b^3) ^(10^-1)
[(a^3 * b^3)^10]^-1
[a^30 *b^30]^-1

Answer 65

10^-1 is not equal to 10*-1

Answer 66

1/10 = 10^-1

Answer 67

true

Answer 68

but that doesn't mean that a^{10^-1} is equal to (a^{10})^{-1}

Answer 69

(a^3 * b^3)^0.1
(a^3*b^3)^1/10
= (a^3*b^3) ^(10^-1)
[(a^3 * b^3)^10]^-1
[a^30 *b^30]^-1
1/[a^30*b^30]
1/a^30 * 1/b^30
a^-30 * b^-30

Answer 70

>.<

Answer 71

yes it does because you can use the rule (ab)^n = a^n*b^n

Answer 72

which we proved for n>0 , for n = 0 it is trivial

Answer 73

0^{-1} is undefined :-P

Answer 74

:D

Answer 75

we assumed that a>0, b>0

Answer 76

right

Answer 77

so it works, but theres a few steps

Answer 78

first we show that (ab)^n = a^n*b^n, for n>0, a>0,b>0 . then we show that
(ab)^(1/n) = a^(1/n)*b^(1/n)

Answer 79

then we can so that (ab)^(p/q) = ...

Answer 80

but still, \[a^{\frac{1}{n}} \neq ((a)^n)^{-1}\]

Answer 81

i didnt write that

Answer 82

you idiot, i wrote

Answer 83

a^(1/n) = [ a^(n)]^-1

Answer 84

you wrote this: = (a^3*b^3) ^(10^-1) [(a^3 * b^3)^10]^-1

Answer 85

and it's wrong :-D

Answer 86

a^(1/n) = a^ [ n^-1] =

Answer 87

no thats right, youre an idiot

Answer 88

lol

Answer 89

(a^3b^3)^(1/10) = (a^3*b^3) ^(10^-1)= [(a^3 * b^3)^10]^-1
get a flipping clue

Answer 90

alright, use the fact that \[a^{n^{-1}} = (a^{n})^{-1} \]to find the square root of 2

Answer 91

because that's what you essentially wrote

Answer 92

no i didnt write that, i have a carat you left out

Answer 93

(a^3*b^3) ^(10^-1)= [(a^3 * b^3)^10]^-1 I didn't leave any carets out

Answer 94

one sec, brb

Answer 95

mine is right, yours is wrong

Answer 96

nope, you've made an error in one of your statements

Answer 97

ok exponentiation is not associative, big whoop

Answer 98

2^(2^3) != (2^2)^3