Recent content by Mike_bb

I wanted to prove that ##y=ln x## without using it as definition. It's not obvious for me that ##y=ln(x)##.

I use definition: ##y=\int_1^x \frac{1}{t}dt##

Hello! Some time ago I had a problem in understanding why ##x = z^y## can't be inverse function of ##y=\int_1^x \frac{1}{t}dt## I decided to prove that ##x=z^y## is not solution of ##y=\int_1^x \frac{1}{t}dt## where ##z## - continuous function that depends on ##y## (##z=f(y)##). Proof: 1...

Perhaps there is mistake in the book: Maybe: ##P=aT'^2+bX'^2+cY'^2+dZ'^2 +....##

Hello! I read about irreducible polynomials and constant factor between them but it was written that polynomials were depended of the same variables. For example, P(x) and Q(x), but in the case of the example we have two polynomial with different variables: P(dt, dx,dy,dz) and...

At the definition stage, we don't know that function ##y=e^x## satisfy to all pair of values ##(x,y)##.

Yes. But how do we know that there is not the pair of values ##(x,y)## for which ##y=e^x## isn't satisfied?

Yes. But that I have putted in my OP is already clear to me. Why can't I ask the second question here?

Not quite. I want to prove that the case ##x \neq e^y## for some ##(X_n;Y_n)## is impossible.

bhobba, Ok. If we define ##log(x)=\int_1^x \frac{1}{t}dt## then it's necessary to prove that ##x=e^y## is really and that ##e## and only ##e## is the base of ##log(x)## i.e. graph of function ##log(x)## with the base ##e## coincides with the graph of function ##y=\int_1^x \frac{1}{t}dt##. How...

How?? I want to define it as ##x=10^y## in such a manner. But it doesn't work. >>It is like force is defined as f=m*a, not suppose f=m*a. I'm not sure I understand what you mean but ##F=ma## is not defined as you want. ##F=ma## has experimental background.

As I understand, we define ##ln(x)## as integral and then suppose that ##x=e^y## and further we find properties of ##e##.

Maybe it's elegant proof, but I like your proof because it shows where ##e^y## came from.

Sagittarius A-Star Big thanks!!! I appreciate you very much! Now I understand how it works!

Sagittarius A-Star In your proof you use the fact that ##exp(x)'=exp(x)##, right?