Mike_bb's latest activity

There mentioned that ##log(10^x)##. But I wrote above ##y=log_{10}(x)## These are two different functions.

As I mentioned above: "I tried to change ##ln(x)## to ##log_{10}(x)## and obtained that ##log_{10}(x)=\int_1^x \frac{1}{t} dt## and its...

That's the point. We get ##log(10^y)##. And as I mentioned in post#1 why do we use ##ln(x)## instead of ##log(10^y)##?

Yes, because you set upper limit of integral as ##e^y## then we would not get ##F(y) = y## after integrating. But we can set upper...

##F'(y)=\frac{1}{10^y}\cdot {d \over dy}(10^y)## Why not?

I think you will see the answer after the calculation (chain rule).

I understand what you mean. But my question is broad: why ##x=e^y## is inverse function instead of ##x=10^y## or ##x=a^y##?

But why? I want to understand why ##x=e^y## is inverse function.

Hello! I have a problem in understanding why ##x=e^y## is inverse function of ##y=\int_1^x \frac{1}{t} dt##. This question seems...

And Years turn into legend. It's a good time to close this thread, which has been spinning around the notion of ##a^0=1## since time...

In math and physics, your IQ means nothing; only the problems you have solved matter.

is wrong, probably you meant: ##(a b)^m = a^m b^m##

So we define ##a^x## such that holds for all real ##n## and ##m##, not just natural numbers. Then we immediately have...

I reread your post. Very interesting and rigorous proof. Big thanks!

1. ##a^m \cdot a^n=a^{m+n}## 2. ##\frac{a^m}{a^n}=a^{m-n}## 3.##(a^m)^n=a^{m \cdot n}## 4.##(ab)^m=a^m \cdot b^m## 5.##\frac {a}{b}^m =...