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

 

[Sagittarius A-Star]

Sagittarius A-Star

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

Here, in the spoiler, I found the same proof, written in a more elegant way:
$$\int\frac1x\,dx,\quad e^u=x,e^u\,du=dx,\,u=\ln(x)$$$$\int\,du=u+C$$$$\int\frac1x\,dx=\ln(x)+C$$

 

[Mike_bb]

Here, in the spoiler, I found the same proof, written in a more elegant way:
$$\int\frac1x\,dx,\quad e^u=x,e^u\,du=dx,\,u=\ln(x)$$$$\int\,du=u+C$$$$\int\frac1x\,dx=\ln(x)+C$$

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

 

[mathwonk]

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?

It helps to have a definition of e. One definition is that e is the unique base such that the derivative of e^y at y=0 equals 1. Since y = 0 corresponds to x =1, this is equivalent to saying that the derivative of its inverse function at x=1 is equal to 1. But the inverse function has derivative 1/x, which does equal 1 at x=1.

 

[bhobba]

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

It is its definition.

You really need to see how all this stuff follows from the definition of ln(x)

We define ln(x), x > 0, called the natural logarithm, as ln(x) = ∫(1 to x) (1/y) dy. Note ln(1) = 0 and ln'(x) = dln(x)/dx = 1/x. ln'(x*y) = y/x*y = 1/x (using f(g(x))’ = df/dx = (df/dg)*(dg/dx) = f'(g(x))*g'(x)). Hence, since they have the same derivative, ln(xy) = ln(x) + C. Let x = 1. ln(y) = C. ln(x*y) = ln(x) + ln(y). This is why the definite integral was from 1 to x. It leads to the simple relation ln(x*y) = ln(x) + ln(y).

An inverse of a function f(x), g(x), is a function such that f(g(x)) = x and g(f(x)) = x. Let e(x) be the inverse of ln(x). That ln(x) has an inverse is evident from its graph. Let a = e(x), b = e(y). ln(a) = x. ln(b) = y. e(x+y) = e(ln(a)+ln(b)) = e(ln(a*b)) = a*b = e(x)*e(y). Let e(1) = e, called Euler’s number. It is a very important number in math. e^n = e*e*…*e, n times. So e^n = e(n). So far, e^n has only been defined for n a natural number. This allows us to define e^x for any x as e(x), and from now on we will use e^x instead of e(x).

a^x is defined as e^(x*ln(a)). a^(x+y) = e^((x+y)*ln(a)) = e^(x*ln(a))*e^(y*ln(a)) = (a^x)*(a^y)

a^x = a^(0 + x) = a^0*a^x. Dividing by a^x, we have a^0 = 1, which is what was asked - it pops out here without any further ado.

a^(x-x) = (a^x)*(a^-x) = 1. a^-x = 1/(a^x)

(a^x)^y = e^(y*ln(a^x)) = e^(y*ln(e^(x*ln(a)) = e^(y*x*ln(a)) = a^(x*y)

a = a^(x/x) = (a^(1 /x))^x or (x√)a = a^(1/x).

From this, you should be able to derive all the exponent rules given in a Precalculus text, only this time for any real numbers. For example x^a/x^b = (x^a)*(1/x^b) = (x^a)*(x^-b) = x^(a-b),

The logarithm of x to base a will be written as loga(x) and is the inverse of a^x, if it exists. Suppose a^x has an inverse, then a^(loga(x)) = x. e^(loga(x)*ln(a)) = x. ln(x) = loga(x)*ln(a). The logarithm to base a is defined as loga(x) = ln(x)/ln(a). We want to show that ln(x)/ln(a) is the inverse of a^x. a^(ln(x)/ln(a)) = e^(ln(a)*ln(x)/ln(a)) = e^ln(x) = x. loga(a^x) = ln(a^x)/ln(a) = ln(e^(x*ln(a))/ln(a) = x*ln(a)/ln(a) = x. Hence the inverse of a^x, loga(x) exists and is ln(x)/ln(a)

loga(x) has the usual properties of logarithms. loga(xy) = ln(xy)/ln(a) = (ln(x) + ln(y))/ln(a) = loga(x) + loga(y).

Now we look at some calculus. log(e^x) = x. (1/e^x)*(e^x)’ = 1. e^x = (e^x)’. The derivative of e^x is the same function. Very interesting.

Let's find the derivative of x^a when a is not just an integer but any real number. (x^a)’ = (e^(a*ln(x))’ = e^(a*ln(x))*(a*ln(x)’ = e^(a*ln(x))*a/x = a*(x^a)/x = a*x^(a-1). Hence, the general formula is the same as when a = n.

Thanks
Bill

 

[Mike_bb]

It is its definition.

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

 

[bhobba]

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

It is defined as the inverse, not a supposition.

It is like force is defined as f=m*a, not suppose f=m*a.

Thanks
Bill

 

[Mike_bb]

It is defined as the inverse, not a supposition.

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.

 

[bhobba]

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.

You may want to define it that way, but that is not its definition in mathematics.

F=M*A is a definition needed for Newton's laws to have physical content, which is, as John Baez expresses it, get thee to the forces.

The same goes for the definition of e^x in math. To conform to its usage in mathematics, it needs to be defined that way. You can't really argue about a definition. The only issue is whether the definition is consistent, and in both cases mentioned, it is.

Thanks
Bill

 

[Matterwave]

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

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