(a^m)^n = a ^(mn)

Harry_the_cat

Harry_the_cat

Elite Member
Joined
Mar 16, 2016
Messages
3,693
\(\displaystyle (a^m)^n=a^{(mn)}\)

For this rule to always apply, we need to specify that a is non-negative. Correct?
 
You are going thorugh the same phase i have gone . In the complex post we discussed (dr p , jeff) about this notation .

Does (a^m)^n Always Equal a^(mn)?

Is the exponent law (a^m)^n = a^(mn) always true? Maybe not! A look at some cases where the law needs some careful interpretation.
www.nctm.org

see this
It will go very complicated belive me .
Also get a idea about multivalued function .
 
See my message on conversation.
I am helping u .I expect the same

It seems like you are also getting confused Which is quite obvious
 
In this post(and the one after) are more detailed constraints for when the "power of power" rule can be applied with non-complex inputs (possibly returning a complex result). As far as I know it's correct :)
 
Cubist said:
In this post(and the one after) are more detailed constraints for when the "power of power" rule can be applied with non-complex inputs (possibly returning a complex result). As far as I know it's correct :)
Click to expand...
4^1/2 is not identical to √4, and in particular the rules for exponents
(a^m)^n = a ^(mn) applied to the former don't work unless you treat it as double-valued.

@Cubist Did u understood
 
Saumyojit said:
4^1/2 is not identical to √4, ...
Click to expand...

4^1/2 = \(\displaystyle \ \dfrac{4^1}{2} \ = \ 2 \ \) because of the Order of Operations.

They (what you have in my quote box) are equal to the same value, but they do
not correspond to each other in meaning of operations.


However, 4^(1/2) corresponds to \(\displaystyle \ \sqrt[2]{4^1}, \ \) and they both equal 2.
 
lookagain said:
4^1/2 = \(\displaystyle \ \dfrac{4^1}{2} \ = \ 2 \ \) because of the Order of Operations.

They (what you have in my quote box) are equal to the same value, but they do
not correspond to each other in meaning of operations.


However, 4^(1/2) corresponds to \(\displaystyle \ \sqrt[2]{4^1}, \ \) and they both equal 2.
Click to expand...
Ok the latter one.
But my point to be clear was not that.
Perhaps see from 25-31 clearly .
U might comment there as my doubts are also not clear .

Complex no

why does -3^2 translate to -9 . I thought -3^2 meant (-3)^2 which gives me 9 . Is there any connection of Bedmas with the evaluation of this expression. First, BEDMAS is a simplifying acronym designed to help beginning students. Second, the minus sign means has at least three different...
www.freemathhelp.com
This post and that have quite similarity.
 
topsquark said:
If m and n are non-zero integers we can have a be less than 0.
Click to expand...

The above is true, but I think it's overly restrictive. The following allows more cases...

If n is a non-zero integer, and m is a non-zero real, we can have a be less than 0.
 
I was reading a article in stack overflow (negative base with fractional expo) where it is told that

Exponention has three meanings as a operator

When we use Negative BASE, then problem arises

Remember ( x^ 1/2 means √x or X ^ 1/3 means cube root....so on )

1: real continuous Exponention operator
Where it is not allowed to have negative bases .
(Normally that we use)

2: discrete real operator which only takes fractional powers whose denominator is only Odd .
As every odd root of negative base gives a real root .
Cube root of -8 is -2 which is real!

3: Expo is a multivalued operator when we are in complex plane or Domain !

where it gives 'n' distinct complex roots ( 'n' is the root value or Denom value of fractional Exponent ^ 1/n)
So cube root of -8 (^1/3 or 3√)
will be {-2, other two conjugate roots}

-----------------------------------------------------------------------
Below is just a analogy that came into my mind which I am sharing

Now I hope you know about this fundamental theorem of algebra .
Where it says
" Every non constant complex polynomial of degree n has exactly n complex roots counted with multiplicity"

Cube root of 8 in complex plane will give me
(Cube root Which now acts as a mutlivalued operator)

Whenever we talk about complex Domain or plane be it any base ( +ve or -ve) the operator either expo and root operator becomes multivalued therefore,

3√8 or (8)^1/3 will give 2 , other two complex conjugate roots .


Now we can also write this in terms of polynomial equation f(X) =x^3- 8
if we want to find roots then f(X)=0
So, x^3-8 = 0
X= 3√8= 2, other two complex conjugate roots.

??
 
Last edited:
You must log in or register to reply here.
Top