##x = z^y## can't be inverse function of ##y=\int_1^x \frac{1}{t}dt##

[Mike_bb]

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. Let $y=\log_z(x)=\int_1^x \frac{1}{t}dt$, then $x=z^y$ (inverse function for $y$).

2. Let $z=f(y)$ and $x=x(y)$ and their inverse functions $y=f^{-1}(z)$ and $y=y(x)$ and $x\ge 1$.

3. From $y=f^{-1}(z)$ and $y=y(x)$ and $x\ge 1$ I conclude that graphs of $y=f^{-1}(z)$ and $y=y(x)$ coincide because $x\ge 1$ and $z>1$.

4. From step 3 it follows that $x=x^y$ has one solution $1=1^0$ but $\log_1(1)$ doesn't exist.

Therefore, if $y=\log_z(x)$ then $z$ can be constant $e$ only and $y=\ln(x)=\int_1^x \frac{1}{t}dt$

Is my proof correct?

Thanks.

 

[anuttarasammyak]

Say $$x=z^y$$, then
$$y=\frac{\ln x}{\ln z}$$. Is it enough ?

Graphs for some z values.

 

[Mike_bb]

Say $$x=z^y$$, then
$$y=\frac{\ln x}{\ln z}$$. Is it enough ?

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

 

[anuttarasammyak]

We can easily integrate your definition integral and get
$$ y=\ln x - \ln 1 = \ln x $$
z does not appear there. So we know yoiur definition gives the condition $$ \ln z =1 $$ to the last equation in post #3 which is a general solution

 

[Mike_bb]

We can easily integrate it and get
$$ y=\ln x $$
z does not appear there. So we know yoiur definition gives the condition $$ \ln z =1 $$ in the last equation in post #3 which is derived without the definitions.

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

 

[Mark44]

Some time ago I had a problem in understanding why $x = z^y$ can't be inverse function of $y=\int_1^x \frac{1}{t}dt$

$x = z^y$ is the inverse of $y = \ln(x)$ if and only if z = e, so $x = z^y$ can be the inverse of $y = \ln(x)$ with this condition on z. Why do you continue writing $\ln(x)$ as a definite integral?

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. Let $y=\log_z(x)=\int_1^x \frac{1}{t}dt$, then $x=z^y$ (inverse function for $y$).

1) You are making this much more difficult than it needs to be, because your equation above involves three unknowns; 2) you seemingly are still confused about logarithm bases; and 3) you complicate things further by writing $\ln(x)$ as a definite integral.

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

Why? Surely you understand that the integral on the right side above is equal to $\ln(x)$, right?

Starting with $y = \ln(x)$, the natural logarithm of x is the exponent on e that results in x, by definition. From the equation just above, y is the exponent on e that results in x. Therefore, $e^y = x$. In short, $y = \ln(x)$ is equivalent to $x = e^y$, making these functions inverses of each other. Again, to find the inverse relationship of an equation that gives y as a function of x, simply solve for x in terms of y. As long as the operations you use are one-to-one, the two equations will be equivalent and the two functions will be inverses. Additionally, the graphs of y = f(x) and $x = f^{-1}(y)$ are identical.

 

[Mark44]

Thread closed. To the OP: It is a violation of forum rules to restart a thread that has been closed.