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

[Mike_bb]

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 strange but I'll try to describe my problem.

As is known, the function $y= \int_1^x \frac{1}{t} dt$ is defined as $y=ln(x)$ but I can't understand why.

In general we can consider the function $y$ as $y=\int_1^x \frac{1}{t} dt$.

At the definition stage, we know nothing about this function (and its properties) but we know only that it has inverse function.

My questions are:

1. If we know that $y=\int_1^x \frac{1}{t} dt$ then why is $x=e^y$ the inverse function of $y$?
2. How to find the inverse function $x=e^y$?

P. S. 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 inverse function is $g(y)=10^y$ :

Let $f(x)=log_{10}(x)$ and $g(y)=10^y$ then $g(y)$ is inverse function to $f(x)$ because: $f(10^y)=log_{10}(10^y)=y$ and $g(log_{10}(x))=10^{log_{10}(x)}=x$

Where did I go wrong?

Thanks.

 

[Sagittarius A-Star]

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

You may differentiate the function $F(y)=\int_1^{e^y}\frac{1}{t}\,dt$ without using the $ln(x)$ function, simplify $F'(y)$ and integrate again.

 

[Mike_bb]

You may differentiate the function $F(y)=\int_1^{e^y}\frac{1}{t}\,dt$ without using the $ln(x)$ function, simplify $F'(y)$ and integrate again.

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

 

[Sagittarius A-Star]

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

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

 

[Mike_bb]

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 $x=a^y$?

 

[Sagittarius A-Star]

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

According to the chain rule:
$F'(y)=\frac{1}{e^y}\cdot {d \over dy}(e^y)$
Maybe you continue the calculation first for $e$ and then repeat the complete calculation for the broader way.

 

[Mike_bb]

According to the chain rule:
$F'(y)=\frac{1}{e^y}\cdot {d \over dy}(e^y)$
Maybe you continue the calculation first for $e$ and then repeat the complete calculation for the broader way.

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

 

[Sagittarius A-Star]

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

Then you would not get $F(y) = y$ after integrating.

 

[Mike_bb]

Then you would not get $F(y) = y$ after integrating.

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 limit of integral as $10^y$ and we get $F(y) = y$.

 

[Sagittarius A-Star]

But we can set upper limit of integral as $10^y$ and we get $F(y) = y$.

That's not true. Proof:
https://www.wolframalpha.com/input?i2d=true&i=Integrate[Divide[1,t],{t,1,Power[10,y]}]

 

[Mike_bb]

That's not true. Proof:
https://www.wolframalpha.com/input?i2d=true&i=Integrate[Divide[1,t],{t,1,Power[10,y]}]

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

 

[Sagittarius A-Star]

The simplification in posting #6 goes:
$F'(y)=\frac{1}{e^y}\cdot {d \over dy}(e^y) = 1$.

 

[Sagittarius A-Star]

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

In Wolfram Alpha they write $log()$ for the natural logarithm, which is usually written as $ln()$.

They mention it:

 

[Mike_bb]

That's not true. Proof:
https://www.wolframalpha.com/input?i2d=true&i=Integrate[Divide[1,t],{t,1,Power[10,y]}]

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 inverse function is $g(y)=10^y$ :

Let $f(x)=log_{10}(x)$ and $g(y)=10^y$ then $g(y)$ is inverse function to $f(x)$ because: $f(10^y)=log_{10}(10^y)=y$ and $g(log_{10}(x))=10^{log_{10}(x)}=x$"

Why can't we define $y=\int_1^x \frac{1}{t}dt$ as $y=log_{10}(x)$ ?

 

[Sagittarius A-Star]

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 inverse function is $g(y)=10^y$ :

Let $f(x)=log_{10}(x)$ and $g(y)=10^y$ then $g(y)$ is inverse function to $f(x)$ because: $f(10^y)=log_{10}(10^y)=y$ and $g(log_{10}(x))=10^{log_{10}(x)}=x$"

Why can't we define $y=\int_1^x \frac{1}{t}dt$ as $y=log_{10}(x)$ ?

Please see above in posting #13.

 

[Mike_bb]

Please see above in posting #13.

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

 

[Sagittarius A-Star]

Why can't we define $y=\int_1^x \frac{1}{t}dt$ as $y=log_{10}(x)$ ?

If you put $10^y$ as upper limit, then you would get $F'(y)=\frac{1}{10^y}\cdot {d \over dy}(10^y) \neq 1$ and $F(y) \neq y$.

 

[Mike_bb]

Sagittarius A-Star

In your proof you use the fact that $exp(x)'=exp(x)$, right?

 

[Mike_bb]

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

 

[Sagittarius A-Star]

you're welcome

 

[Jean-Claude]

Seems to me the error is in the sentence before "P.S." end ":"
precisely just after "log10(x)="
A coef is missing