# Math Help - Why is log(0,0)=indeterminate while 0^1=0? Why is log(0,5)=0 if 0^0=indeterminate?

1. ## Why is log(0,0)=indeterminate while 0^1=0? Why is log(0,5)=0 if 0^0=indeterminate?

Hello everyone.
I'm wondering why log(0,0) is indeterminate while 0^1=0?
And why log(0,5)=0 exponent when 0^0=indeterminate?
I'm a little bit confused here.
Thanks.

2. ## Re: Why is log(0,0)=indeterminate while 0^1=0? Why is log(0,5)=0 if 0^0=indeterminate

Originally Posted by BrokenCalculator
Hello everyone.
I'm wondering why log(0,0) is indeterminate while 0^1=0?
And why log(0,5)=0 exponent when 0^0=indeterminate?
What you posted makes no sense.
Give an example if necessary.

3. ## Re: Why is log(0,0)=indeterminate while 0^1=0? Why is log(0,5)=0 if 0^0=indeterminate

log(0,0) isn't "indeterminant", it is "undefined". I don't know what 0^1= 0 has to do with that.
I don't know what you mean by "log(0,5)= 0 exponent". Assuming you mean the natural logarithm, log(0,5)= -0.693, approximately, not 0.
0^0 is indeterminant because the are different function which if you replace the variable by 0, give "0^0" but have different limits as the variable goes to 0. One example is 0^x as x goes to 0, which has limit 0 as x goes to 0, another is x^0 as x goes to 0, which has limit 1 as x goes to 0.

4. ## Re: Why is log(0,0)=indeterminate while 0^1=0? Why is log(0,5)=0 if 0^0=indeterminate

Originally Posted by Plato
What you posted makes no sense.
Give an example if necessary.
I've always thought logarithm operation was opposite of exponentiation just like division is opposite of multiplication.
Here is an example to show what I mean:
3^4=81 or 1*3*3*3*3=81
Now, log(3,81)=4 or 81/3/3/3/3=1

Wouldn't log(0,0) be equal to 1 considering an example above and since 0^1=0?
The same goes for log(0,5), how can it be 0 if 0^0=indeterminate?

Originally Posted by HallsofIvy
log(0,0) isn't "indeterminant", it is "undefined". I don't know what 0^1= 0 has to do with that.
I don't know what you mean by "log(0,5)= 0 exponent". Assuming you mean the natural logarithm, log(0,5)= -0.693, approximately, not 0.
0^0 is indeterminant because the are different function which if you replace the variable by 0, give "0^0" but have different limits as the variable goes to 0. One example is 0^x as x goes to 0, which has limit 0 as x goes to 0, another is x^0 as x goes to 0, which has limit 1 as x goes to 0.
Sorry, my Microsoft Math shows the result as indeterminate, so I decided to call it that.
As far as my understanding goes, 0^1=0 shows the exponent, or the resultant value of log(0,0) should be 1. Isn't logarithm used to find exponents? Why doens't it match?
May I ask what you mean by "natural logarithm"? I'm new to math and unfortunately don't know much. Please guide me in the right direction with a short explanation if you will, what natural exponents are and what other exponents there are so I can start learning what's needed.
Thanks.

5. ## Re: Why is log(0,0)=indeterminate while 0^1=0? Why is log(0,5)=0 if 0^0=indeterminate

you are confusing the BASE with the EXPONENT.

log(0) would be: "what number is it such that e^x = 0?" there is no such number x. if there WAS such a number, it would be more negative than any other number. we could (and some do) call:

log(0) = -∞.

the opposite OPERATION of "exponentiation" (for a fixed exponent a), that is to say x^a, is taking the a-th ROOT, that is: x^(1/a).

if you are looking for something like: "log base 0", that doesn't work. the function 0^x is 0 at every number x except 1 (a horizontal line with perhaps one point missing, or moved). the "inverse function" of that would be a VERTICAL line x= 0, which makes it impossible to determine WHICH y-value it takes at x = 0, and there ISN'T any y-value at other x's.