Thread: Is this allowed?

Hence, 0=1 ???

What rule have I broken, or not obeyed?? Surely this isn't right lol.

Works for higher powers too. So 0=everything? Whats the implication???

Originally Posted by sox

Works for higher powers too. So 0=everything? Whats the implication???

You are misapplying the logarithm.

It doesn't work for higher powers for complex numbers either -- think about Euler's formula.

Also, I believe that, to reach that solution, you would have to do a division by 0, which is not allowed, my friend. :wink: . Also, if no division by 0 occurs, I conclude that you would get 0=0 for an answer.

Originally Posted by sox

Works for higher powers too. So 0=everything? Whats the implication???

Again, only works if you divide by zero.

Let f(x)=1^x. This function is independent of x. As someone has observed, you can take logs to get xlog(1) which is independent of x. However log(1)=0, so you can't conclude anything.

1 to the power 0 is basically 1 X 0

anything times 0 will always = 0 (because there are zero copies of the number)

0=0 because multiplying any number by zero, is really just multiplying zero by zero

Anything to the power 0 is 1, not 0. (Edit: except 0^0, which is undefined.)

Originally Posted by sox

Hence, 0=1 ???

What rule have I broken, or not obeyed?? Surely this isn't right lol.

You're claiming that if , then x = y.

That would be true for positive since, after taking logarithms of both sides, you'd get

, and therefore x = y. But if a=1, log(a) =0, so you can have

for any x,y.