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.
Looking forward for your help.
Thanks.
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.
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?
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.
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.