How can we prove ##e^{ln x}= x## and ##e^-{ln(x+1)}= \frac 1 {x+1}##?

[chwala]

Homework Statement

they say 1. $e^{ln x}= x$ and 2.$e^-{ln(x+1)}= \frac 1 {x+1}$ how can we prove this $e^{ln x}= x$ and also $e^-{ln(x+1)}= \frac 1 {x+1}$?

Homework Equations

The Attempt at a Solution

let $ln x = a$ then
$e^a= x,$ a ln e= x,$→a= x, where$ ln x= x

 

[Delta2]

Set $y=ln(x+1)$ so we ll have that $e^y=x+1$.

 

[chwala]

Delta what are you doing? so?

 

[Delta2]

Sorry I now see you have edited the original post.

It is from the definition of the natural logarithm $lnx$ that $e^{lnx}=x$. So not much to prove here, it follows directly from the definition.

Now we want to look at $e^{-ln(x+1)}$. This is equal to $\frac{1}{e^{ln(x+1)}}$ don't you agree? (it is like saying that $e^{-y}=\frac{1}{e^y})$

 

[chwala]

Thanks a lot, i was blind but now i see
let $ln x = y$...1
it follows that $e^y = x$
taking logs on both sides,
$y ln.e = ln x$
$y = ln x$ now substituting this in original equation 1,
$ln x = ln x$,
implying that $e^{ln x}= x$
greetings from Africa, chikhabi

 

[Delta2]

This is correct but what you doing is like driving in a circle, i.e you start from $lnx=y$ to end up with $y=lnx$ which is essentially the same thing.

Substituting $y=lnx$ to $e^y=x$ is what proves (if we can call this a mini proof) that $e^{lnx}=x$.

Greetings from Athens, Greece.