Why is ##x=e^y## the inverse of ##y=\int_1^x \frac{1}{t} dt##?

[MidgetDwarf]

If you have access to a library or even Google (cough pdf cough)

I would find Moise Calculus. He has a section as mathwonk stated, starting with a definition for e.

If the book is to hard to find (a physical copy is expensive and hard to track down), then look for Stewart calculus. In the 7th edition, he has optional section (optional sections are *), in which he starts with ln to get e^x.

Its been a while since I looked at Stewart(10yrs), but I'm sure he does this .

 

[Mark44]

Now we look at some calculus. log(e^x) = x.

I think you meant to write $\ln(e^x) = x$. The symbol log is commonly, but not universally, taken to mean $\log_{10}$. In computer science textbooks, the symbol log usually means $\log_2$.

How?? I want to define it as $x=10^y$ in such a manner. But it doesn't work.

The inverse relationship to the above is $y = \log_{10} x$. If it is understood that we're talking about the common log (log base 10), $y = \log x$.

What is the "it" that you want to define as $x=10^y$?

See above.

 

[Mark44]

From post #1:

I have a problem in understanding why $x=e^y$ is inverse function of $y=\int_1^x \frac{1}{t} dt$.

There are several posts in this thread that seem to overlook the simplest way to explain things.

Let's look at the second function above.
$y=\int_1^x \frac{1}{t} dt = \ln(t)|_1^x = \ln(x) - \ln(1) = \ln(x)$. So a simpler form of the equation above on the right is $y = \ln(x)$.

In general, to find the inverse of a function x = f(y), where x is given in terms of y, you simply solve the equation for y in terms of x. As long as whatever operations you apply are one-to-one, and it is possible to solve the starting equation, you will end up with the inverse function. This will be in the form of $y = f^{-1}(x)$.

Starting from $x = e^y$, take the natural log of both sides. I.e., $\ln(x) = \ln(e^y) = y$. IOW, $y = \ln(x)$.

Note that some operations used to solve an equation are not one-to-one, so you might not end up with an actual function. For example, if $y = x^2$, I can solve for x in terms of y by taking the square root of both sides. It's important to note that positive numbers have two square roots, so taking the square root of both sides is not a one-to-one operation. So $\sqrt{y} = |x|$. Equivalently, $x = \pm\sqrt {y}$, so we ended with a relation that is not a function.

There also seems to be some confusion on the part of the OP here about which log functions are inverses of which exponential functions. In general $\log_b(b^x) = x$, but $\log_b(a^x) \ne x$.
Some examples:
$\log_{10}(10^x) = x$ (This is the same as $\log(10^x) = x$ with the understanding that $\log$ means $\log_{10}$).
$\log_e(e^y) = y$
$\log_2(2^z) = z$

 

[Mike_bb]

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 to prove it?
Thanks.

 

[PeroK]

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 to prove it?
Thanks.

That is precisely the question in your OP. I would forget $\log$ or $\ln$. You have an integral function and the defining property of the exponential is that $\dfrac{d}{dy} e^y = e^y$.

You need to show these are inverse functions.

Hint: use an integral substitution.

 

[Mike_bb]

That is precisely the question in your OP.

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

 

[PeroK]

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

That is definitely not what you put in your OP!!

 

[Mike_bb]

That is definitely not what you put in your OP!!

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

 

[bhobba]

then it's necessary to prove that $x=e^y$ is really

What the above statement means has me beat. By definition, e^x is the inverse of ln(x); ln(x) is only defined where x > 0. The only issue is that ln(x) is invertible, so the definition makes sense. The derivative of ln(x) is 1/x, which is strictly decreasing, so when you graph it, it must have an inverse (think about it - in a real analysis course you would prove it rigorously, but in plain calculus you can be less strict and appeal to geometric intuition).

Thanks
Bill

 

[Mike_bb]

By definition, e^x is the inverse of ln(x).

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

 

[Mike_bb]

What the above statement means has me beat.

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

 

[bhobba]

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

We do. It is implied by the definition. When one makes a statement, all its implications are immediately true; they are just made explicit through further analysis, but remain true regardless of whether or not that analysis is done.

Thanks
Bill

 

[PeterDonis]

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

If you define $e^x$ as the inverse of $\ln x$, you are making use of the definition of the term "inverse", which says that, given a function $f$ that is a one-to-one, onto mapping of a domain $D$ to a range $R$, the inverse $f^{-1}$ is a one-to-one, onto mapping of $R$ to $D$.

For this case, we have $f = \ln x$, $D$ is the set of positive real numbers, $R$ is the entire set of real numbers, and the notation $e^x$ is $f^{-1}$ by definition. That establishes that for every pair of values $(x, y)$ for which $y = \ln x$, it is also true that $x = e^y$. So we do know "at the definition stage" the thing you're concerned about.

 

[bhobba]

Peter's answer is exact, and IMHO leaves nothing left to be said.

Thread closed.

Thanks
Bill