relation betwn ln and e

Question

If $f(x) = ln(x)$ and $f^-1(x) = e^x$ then is $e^x = 1/ln(x)$???

because I see $e^9 = 8103$ but $1/ln9 = .455$

How are they reverse? I don't understand!

Answer 1

$f^{-1}$ is the inverse function of $f$. That is, $$ f(f^{-1}(x)) = x, f^{-1}(f(x)) = x $$ when this statement makes sense. It is not $\frac{1}{f}$ or anything. The $^{-1}$ is not an exponent.

Answer 2

As copper.hat mentioned it, there is a notation "clash" here.

More precisely, let there be $E$ a set and $$\matrix{\star: & E^2 & \to & E\\ & a,b & \mapsto & \star(a, b) & =a \star b}$$

such that $(E, \star)$ is a group.

• Either we notate $\star$ like the addition and name $0_E$ the neutral element of E, and $\forall a \in E, ~~-a$ its symmetrical, so that $a\star(-a) = (-a)\star a = 0_E$
• Or we can notate $\star$ like the multiplication and name $1_E$ the neutral element, and $\forall a \in E ~~a^{-1}$ its symmetrical. Here again, $a \star a^{-1} = a^{-1} \star a = 1_E$

It actually makes no difference, but can be useful when manipulating several laws, for instance in rings, fields or *-Algebra.

Now back to the question

The field we use the most is $(\mathbb{R}, +,\times)$ So in most situations we encounter, when seeing something like $-a, a^{-1}$, assuming they are the (respectively) opposite and inverse of a real named $a$ is not that risky.

But the set of all bijections from $E$ to $F$, $A$, comes with a defined law, noted $\circ$ such that : $$\matrix{\circ : & A^2 & \to & A\\ & u,v & \mapsto &\circ(u,v) & = u\circ v : & E & \to & F\\ &&&& & x & \mapsto & (u\circ v)(x) & = u(v(x)) }$$

And, as $(A,\circ)$ turns out to be a group, any bijection of A has a unique symmetrical in regards of $\circ$.

And as we notate $\circ$ just like multiplication, we notate it $f^{-1}, \forall f \in A$, such that $f\circ f^{-1} = f^{-1}\circ f = 1_A$.

note : if $(E = F), 1_A = id_E : x \mapsto x$

Problem

let $f \in A, x\in E$,

And that's were it get confusing : we know that $f(x)\in F$, so if $(F,\times)$ is a group, when we read $(f(x))^{-1}$, sometimes noted $f^{-1}x$ not to confuse with $fx^{-1}$, we may safely assume that it's the inverse of $f(x)$ in regards of the multiplication defined on $F$.

But now if we have $y \in F, f^{-1}(y)$ now refers to the image of $y$ through $f^{-1} : F \to E$ the inverse function of $f$.

But what now if $E=F=\mathbb{R}$ ? well, being rigorous is the only way.

Solution

First right down the exact sets where your functions are defined. For instance : $$ \matrix{exp : & \mathbb{R} & \to & \mathbb{R}^{+*}\\ln : & \mathbb{R}^{+*} & \to & \mathbb{R}}$$ You first notice that the output of one is included or equal to the input of the other...

Answer 3

First, you have to know, that if there's an little $-1$ over a function like $f^{-1}(x)$ that doesn't mean that it equals $\frac1{f(x)}$ at all.

• $f^{-1}(x)$ is a way to tell that the function $f^{-1}(x)$ is the inverse of the function $f(x)$. So when we input some number into a function, we get an output. If we input the output into the inverse function, you will get number you inputed at first.

• For example, let's say we have a function $f(x)=2x+1$. The inverse of this function would be $f^{-1}(x)=\frac{x-1}2$. For example if you plug in $3$ for $x$, you will get $f(3)=2(3)+1=7$. Now, you plug in the answer to get you back to the starting number: $f^{-1}(7)=\frac{7-1}2=\frac62=3$.

• As you can see from the example above, the functions cancel each other out, so if we plugin some $x$, and do either $f^{-1}(f(x))$ or $f(f^{-1}(x))$, you will get the number you started with, $x$. This works for any function $f(x)$ and its inverse.

• As you probably recall, $\log_ex$ or $\ln x$ is an inverse function of $e^x$. So from the property $f(f^{-1}(x))$, we know that if we do $\ln e^x$ or $e^{ln \,x}$ wou will get $x$, the number you started with.

• So if you input $e^9$, you will get answer of $8013.0839...$ Now if you plug in the answer into a natural logarithm (the inverse of exponential function), you will get $\ln8013.0839...=9$.

Answer 4

To avoid any ambiguity between the reciprocal and the inverse, one may write $$f^{-1}$$ for the reciprocal and $$\mathop f^{-1}$$ for the inverse.