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## #1 2014-04-02 08:07:24

knightstar
Member
Registered: 2014-03-12
Posts: 14

### Inverse Operations

I have a couple questions in regards to the relationship and evidently non-relationship between exponentiation, logarithms and nth roots.
For starters let me just start of at the beginning...

...Addition is the operation of combining quantities
Subtraction is the operation of inverting addition...

...Multiplication is the operation of repeating addition
Division is the operation of inverting multiplication...

Exponentiation is the operation of repeating multiplication...

...but what is the inverse operation of exponentiation? I read in wiki that the inverse of exponents are logarithms and if this is true, then what about nth roots?
All three of these terms seem to be directly related, even though nth root is given different terms for its inputs and output.

It's true that...
Base is to exponentiation (and logarithm) as root is to nth root
Exponent is to exponentiation (and logarithm) as index is to nth root
Power is to exponentiation (and logarithm) as radicand is to nth root

E.G.
2^3= 8 as the LOG base 2 of 8= 3 as the cubed root of 8= 2

So if logarithm is in fact the inverse of exponentiation, then why? And what is the inverse operation of nth root?

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## #2 2014-04-02 08:16:32

ShivamS
Member
Registered: 2011-02-07
Posts: 3,648

### Re: Inverse Operations

Here's a good article: http://www.thejach.com/view/2010/1/rooting_is_not_the_inverse_of_exponentiating
It's a bit contradicting. Some people say both taking a root and log are the inverses.

Last edited by ShivamS (2014-04-02 08:17:35)

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## #3 2014-04-02 08:37:04

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Inverse Operations

I'd say both are true, it just depends on what you take as your variable.

The inverse of
is
.

The inverse of

is
.

Last edited by anonimnystefy (2014-04-02 08:37:40)

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #4 2014-04-02 12:15:21

eigenguy
Member
Registered: 2014-03-18
Posts: 78

### Re: Inverse Operations

Exponentiation is a binary operation. You can think of it as a function, but it is not 1-1, so it does not have an inverse in and of it self.

By fixing one of the operands, exponentiation can define two functions, both of which are locally 1-1, and thus do have local inverses:

Power functions fix the exponent and allow the base to vary:  y = x[sup]c[/sup]. The inverse of this function is taking the root = power function with multiplicative inverse for the exponent.

Exponential functions fix the base and allow the exponent to vary:  y = c[sup]x[/sup]. The inverse of the function is the corresponding logarithm function.

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich

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## #5 2014-04-04 07:52:06

knightstar
Member
Registered: 2014-03-12
Posts: 14

### Re: Inverse Operations

I met up with my math tutor today and he wasn't sure, but he suggested that perhaps factorization was the inverse operation of exponentiation.

That is, if exponentiation is repeated multiplication, then factorization would be repeated division.

For example 2x2x2= 8 as 8/2/2= 2.

He also pointed out that as ShivamS suggested, in a way, nth root and logarithm are inverse operations of eachother.

That is, the index of the nth root is the solution (or exponent) of logarithm, and the base of logarithm is the solution (or root) of nth root.

For example, the 3rd root of 8= 2 as the log base 2 of 8= 3.

If this is a practical understanding, I wonder now why there is little on the internet to support it. So what do you all think about this? Can factorization be considered an operation?

How could you define nth root and logarithm in relation to exponentiation and factorization, as they are in relation to multiplication and division, and as they are in relation to addition and subtraction?

It all seems very fitting so far, though of course I'm not so far along as you all. Would this understanding become a problem later on as my mathematical education advances?

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## #6 2014-04-04 07:54:56

ShivamS
Member
Registered: 2011-02-07
Posts: 3,648

### Re: Inverse Operations

I never said the nth root and logarithm are inverse operations of each other. I said that certain people believe that they both are the inverse operations of exponentiation. This will not be a problem for you, ever. Many parts of math are not well defined, so a few different opinions won't hurt your understanding of the subject.

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## #7 2014-04-04 08:31:17

knightstar
Member
Registered: 2014-03-12
Posts: 14

### Re: Inverse Operations

I just want to make sure my understanding of certain fundamental terms will not ultimately be contradictorial. I apologize for my misinterpretation.

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## #8 2014-04-04 08:32:31

ShivamS
Member
Registered: 2011-02-07
Posts: 3,648

### Re: Inverse Operations

How are you thinking of exponentiation? x^n as a function of x for fixed n? Or a^x for fixed a? If the former then the inverse function is the nth root. If the latter then it is the logarithm base a.

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