The logarithm and exponential functions

The logarithm functions\(: ln(x), log_a(x)\)

Both functions \(log_a(x)\) and \(ln(x)\) being defined as similar up to a coefficient \(ln(a)\), the following properties will be true for both logarithmic functions.

Definition of the natural logarithm

$$ \forall x \in \mathbb{R^+}, $$

$$ ln(x) = \int^x_1 \frac{dt}{t} $$

As well we do have:

Logarithm of a product \(: ln(ab)\)

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ ln(ab) = ln(a)+ ln(b) $$

Logarithm of an inverse\(: ln \left(\frac{1}{a} \right)\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ ln \left(\frac{1}{a} \right) = -ln(a) $$

Logarithm of a quotient\(: ln \left(\frac{a}{b} \right)\)

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ ln \left(\frac{a}{b} \right) = ln(a) - ln(b) $$

Logarithm of a power\(: ln \left( a^n \right)\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall n \in \hspace{0.05em} \mathbb{R},$$

$$ ln(a^n) = n \ ln(a) $$

Link between logarithm to the base a or b and natural logarithm\( : log_a(x), log_b(x), ln(x)\)

In the general way, for two logarithms respectively to the base \(a\), \(b\) or \(e\) (natural logarithm),

The exponential functions\(: e^x, a^x \)

Exponential functions, being defined as the reciprocal function of the logarithm (respectively to its base), have all the reciprocal properties of logarithms.

The definition of the exponential function

Product of an exponential

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ e^a \ e^b = e^{a+b}$$

Inverse of an exponential

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ \frac{1}{e^b} = e^{- b}$$

Quotient of an exponential

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ \frac{e^a}{e^b} = e^{a - b}$$

Exponential to the power of n

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ (e^a)^b = e^{ab} $$

Value of \(e\)

$$ $$

$$ e \approx 2.7182818... $$

Link between powers to the base a or b and the exponential function\(: a^x, b^x, e^x\)

Recap table of logarithm and exponential functions

---

Demonstrations

The logarithm functions\(: ln(x), log_a(x)\)

Definition of the natural logarithm

Let \(a \in \mathbb{R}\) be a real number and \(f\) a function such as:

$$\forall x \in \mathbb{R}, \ f : x \longmapsto a^x$$

$$ \hspace{5.5em} \mathbb{R} \longmapsto \mathbb{R^+}$$

The origin oh the logarithm comes from the question of determining a function that returns the exponent \(x\) of a function \(f(x) = a^x\). Therefore, this function will be the reciprocal function of the function \(f\), because it returns identity \((f \circ f^{-1})\):

$$f^{-1} : a^x \longmapsto x $$

$$ \hspace{2.6em} \mathbb{R^+} \longmapsto \mathbb{R}$$

So,

$$ ( f^{-1} )'= \frac{1}{ ((a^x)'\circ f^{-1})} \qquad (1) $$

But, by the definition of a derivative:

$$ (a^x)' = lim_{h \to 0 } \enspace \frac{a^{x + h} - a^x}{h} $$

$$ (a^x)' = lim_{h \to 0 } \enspace \frac{a^x a^h - a^x}{h} $$

$$ (a^x)' = a^x \ lim_{h \to 0 } \enspace \frac{a^h - 1}{h} $$

If we take the hypothesis that this limit exists, and we call it \(\alpha\), we then have:

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \ (a^x)' = \alpha \ a^x \qquad (2) $$

Now injecting the expression \((2)\) into \((1)\), we obtain:

$$ ( f^{-1} )'= \frac{1}{ (\alpha \ a^x \circ f^{-1})} $$

$$ ( f^{-1} )'= \frac{1}{ \alpha ( a^x \circ f^{-1})} $$

But, \(a^x = f\), so:

$$ ( f^{-1} )'= \frac{1}{ \alpha x} $$

Now, taking the antiderivative on each side of the equation:

$$ \int^x ( f^{-1}(t) )' \ dt = \int^x \frac{1}{ \alpha t} \ dt $$

We thus can define this function by an integral:

$$ \int^x_1 ( f^{-1}(t) )' \ dt = \int^x_1 \frac{1}{ \alpha t} \ dt $$

$$ f^{-1}(x) - \ \underbrace{ f^{-1}(1) } _\text{\( = \ 0 \)} \ = \frac{1}{ \alpha } \int^x_1 \frac{1}{t} \ dt $$

$$ f^{-1}(x) = \frac{1}{ \alpha } \int^x_1 \frac{dt}{t} \qquad (3) $$

We no have a definition of what is the Napierian logarithm (or natural logarithm):

$$ \forall x \in \hspace{0.05em} \mathbb{R}^+, $$

$$ ln(x) = \int^x_1 \frac{dt}{t} $$

Consequently, we do also have:

$$ ln(1) = \int^1_1 \frac{dt}{t} = 0$$

$$ ln(1) = 0 $$

Moreover:

$$ \forall x \in \hspace{0.05em} \mathbb{R}^+, \ ln(x)' = \ \left( \int^x_1 \frac{dt}{t} \right)' $$

$$ ln(x)' = \frac{1}{x} $$

So,

$$ \forall x \in \hspace{0.05em} \mathbb{R}^+, $$

$$ ln(x)' = \frac{1}{x} $$

In the end, with the expression \((3)\), and for all power of \(a\):

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \ log_a(x) = \frac{ln(x) }{\alpha} \qquad (4) $$

with a certain parameter \(\alpha\) still to determine.

Both \(log_a(x)\) and \(ln(x)\) functions being defined as the same thing up to a coefficient \(\alpha\), the following properties will be true for both logarithmic functions.

Logarithm of a product \(: ln(ab)\)

Using the formula found previously, we have:

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ ln(ab) = \int^{ab}_{1} \frac{dt}{t}$$

Now using the Chasles relation applied to integrals:

$$ ln(ab) = \int^{a}_{1} \frac{dt}{t} + \int^{ab}_{a} \frac{dt}{t}$$

By setting down a variable change for the right integral:

$$ \Biggl \{ \begin{gather*} t = a \ u \\ dt = a \ du \end{gather*} $$

$$ ln(ab) = ln(a) + \int^{b}_{1} \frac{ a \ du }{a \ u } $$

$$ ln(ab) = ln(a) + \int^{b}_{1} \frac{ du }{ u } $$

And as a result,

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ ln(ab) = ln(a) + ln(b) $$

Logarithm of an inverse\(: ln \left(\frac{1}{a} \right)\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+,$$

$$ ln \left(\frac{1}{a} \right) = \int^{\frac{1}{a}}_{1} \frac{dt}{t}$$

By setting down a variable change for this integral:

$$ \Biggl \{ \begin{gather*} t = \frac{u}{a} \\ dt = \frac{du}{a} \end{gather*} $$

$$ ln \left(\frac{1}{a} \right) = \int^{1}_{a} \frac{du}{a} \frac{a}{u}$$

$$ ln \left(\frac{1}{a} \right) = \int^{1}_{a} \frac{du}{u} $$

$$ ln \left(\frac{1}{a} \right) = -\int^{a}_{1} \frac{du}{u} $$

And finally,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ ln \left(\frac{1}{a} \right) = -ln(a) $$

Logarithm of a quotient\(: ln \left(\frac{a}{b} \right)\)

By simply taking the two previous properties, we have:

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ ln \left(\frac{a}{b} \right) = ln \left(a \times \frac{1}{b} \right) $$

$$ ln \left(\frac{a}{b} \right) = ln(a) - ln \left(\frac{1}{b} \right) $$

And as a result,

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ ln \left(\frac{a}{b} \right) = ln(a) - ln(b) $$

Logarithm of a power\(: ln \left( a^n \right)\)

By retrieving the previous relation:

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2,$$

$$ ln(ab) = ln(a) + ln(b) $$

And replacing \(b\) by \(a\), we obtain:

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ ln(a^2) = ln(a) + ln(a) $$

$$ ln(a^2) = 2 \ ln(a) $$

Now calculating \(ln(a^3)\):

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+,$$

$$ ln(a^3) = ln(a^2 \times a) $$

$$ ln(a^3) = ln(a^2)+ ln(a) $$

$$ ln(a^3) = 2 \ ln(a) + ln(a) $$

$$ ln(a^3) = 3 \ ln(a) $$

We see that by a direct recurrence, we obtain:

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall n \in \hspace{0.05em} \mathbb{N}, $$

$$ ln(a^n) = n \ ln(a) $$

(idem moving backwards for \( n \in \hspace{0.05em} \mathbb{Z}\))

Furthermore, if we now take \( n \in \hspace{0.05em} \mathbb{R}\), we are forced to try with another method:

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall n \in \hspace{0.05em} \mathbb{R}, $$

$$ ln(a^n) = \int^{a^n}_{1} \frac{dt}{t}$$

By setting down the variable change:

$$ \Biggl \{ \begin{gather*} t = u^n \\ dt = n \times u^{n-1} \ du \end{gather*} $$

we then obtain:

$$ ln(a^n) = \int^{a}_{1} \frac{ n \times u^{n-1} }{u^n} \ du $$

$$ ln(a^n) = \int^{a}_{1} \frac{ n \times u^{n-1} }{u^{n-1} \times u} \ du $$

$$ ln(a^n) = n \int^{a}_{1} \frac{du}{u} $$

And as a result,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall n \in \hspace{0.05em} \mathbb{R},$$

$$ ln(a^n) = n \ ln(a) $$

Link between logarithm to the base a or b and natural logarithm\( : log_a(x), log_b(x), ln(x)\)

Retrieving the previous found expression:

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \ log_a(x) = \frac{ln(x) }{\alpha} \qquad (4) $$

The logarithm to the base \(a\), which is the reciprocal function of the starting function \(a^x\), is defined up to a coefficient from the natural one. Let us study the behaviour og this function for \(x = a^n\):

$$ log_a( a^n) = \frac{ln( a^n) }{\alpha} \qquad (4) $$

$$ n = n \ \frac{ln(a)}{\alpha} $$

$$ \alpha = ln(a) $$

So, having found the coefficient between both logarithms:

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ log_a(x) = \frac{ln(x) }{ln(a)} $$

And in this cas, its derivative is worth:

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ log_a(x)' = \left( \frac{ln(x) }{ln(a)}\right)' $$

$$ log_a(x)' = \frac{ln(x)'}{ln(a)} $$

So,

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ log_a(x)' = \frac{1}{x \ ln(a)} $$

In the general case, for two logarithms respectively to the base \(a\) and \(b\), we study:

$$ \Biggl \{ \begin{gather*} log_a(a^x) = x \\ log_b(a^x) = x \ log_b(a) \end{gather*} $$

Then, combining both expressions we do quickly find that:

$$log_b(a^x) = log_a(a^x) \ log_b(a) $$

And as a result,

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall b \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ log_a(x) = \frac{ log_b(x) }{log_b(a)} $$

Furthermore, the previous relation and its twin lead us to:

$$ \forall (a, b) \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr]^2, \ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, $$

So,

$$ \forall (a, b) \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr]^2, $$

$$ log_a(b) = \frac{1}{log_b(a)} $$

The exponential functions\(: e^x, a^x \)

The definition of the exponential function

Now consider the reciprocal function of the function \(g(x) = ln(x)\).

If we take the formula for the derivative of a reciprocal function, we do have:

$$ \forall (g,g^{-1}) \in F(\mathbb{R}, \mathbb{R})^2, \enspace (g' \circ g^{-1}) \neq 0, $$

$$ ( g^{-1} )'= \frac{1}{ (g' \circ g^{-1})} $$

So,

$$ ( g^{-1} )'= \frac{1}{ (ln(x)' \circ g^{-1})} $$

$$ ( g^{-1} )'= \frac{1}{ \frac{1}{g^{-1}} } $$

$$ ( g^{-1} )'= g^{-1} $$

This function has itself for derivative !

It is the exponentiel function which we will note \(e^x\) but sometimes \(exp(x)\). Indeed, we will later on that it has all the properties of a power function of x.

Product of an exponential

1. Using the reciprocal properties of the logarithm

Having said that \(g(x) = ln(x)\) has the following property:

$$xy \longmapsto ln(xy) = ln(x) + ln(y) $$

$$ \left(\left[ \mathbb{R^*_+} \right]^2 \longmapsto \hspace{0.05em} \mathbb{R}^2\right)$$

If we set down new variables:

$$ \Biggl \{ \begin{gather*} X = ln(x) \Longleftrightarrow x = e^{X}\\ Y = ln(y) \Longleftrightarrow y = e^{Y} \end{gather*} \qquad (5) $$

Then, the reciprocal function \(g^{-1} = e^x\) must have this property:

$$ ln(xy) = X + Y \longmapsto xy = e^{X + Y} \qquad (6)$$

$$ \left(\mathbb{R}^2 \hspace{0.05em} \longmapsto \hspace{0.05em}\left[ \mathbb{R^*_+} \right]^2\right)$$

Both expressions \((5)\) and \((6)\) together show that:

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ e^a \ e^b = e^{a+b}$$



2. Using a Taylor series

We know that the Taylor series of the exponential function at zero is:

$$ \forall x \in \mathbb{R}, \ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ ... \ + \frac{x^n}{n!}$$

Moreover, using the general formula of Taylor series at \(a\) under the form:

$$ f(a + b) = f(a) + f'(a)b + \frac{f^{(2)}(a)}{2!}b^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}b^n $$

But, in the case of the exponential function, we had this property:

$$ \forall x \in \mathbb{R}, \ (e^x)' = e^x $$

Which implies that,

$$ f(a) = f'(a) = f^{(2)}(a) = \ ... \ = f^{(n)}(a) $$

So now,

$$ f(a + b) = f(a) + f(a)b + \frac{f(a)}{2!}b^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f(a)}{n!}b^n $$

$$ f(a + b) = f(a) \underbrace{ \left(1 + b + \frac{b^2}{2!} + \frac{b^3}{3!} + \ ... \ + \frac{b^n}{n!} \right) } _\text{\( = \ f(b)\)}$$

$$ f(a + b) = f(a) f(b)$$

And as a result,

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ e^a \ e^b = e^{a+b}$$

Inverse of an exponential

Using the same logic as above,

$$ \frac{1}{y}\longmapsto ln\left(\frac{1}{y}\right) = - ln(y) $$

$$ \left(\mathbb{R^*_+} \longmapsto \hspace{0.05em} \mathbb{R}\right)$$

If we set down,

$$ Y = ln(y) \Longleftrightarrow y = e^{Y} \qquad (5) $$

So,

$$ ln \left(\frac{1}{y} \right) = - Y \longmapsto \frac{1}{y} = e^{- Y} \qquad (7)$$

$$ \left(\mathbb{R} \hspace{0.05em} \longmapsto \hspace{0.05em} \mathbb{R^*_+}\right)$$

Both expressions \((5)\) and \((7)\) together show that:

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ \frac{1}{e^b} = e^{- b}$$

Quotient of an exponential

Using the same logic as above,

$$ \frac{x}{y} \longmapsto ln\left(\frac{x}{y}\right) = ln(x) - ln(y) $$

$$ \left(\left[ \mathbb{R^*_+} \right]^2 \longmapsto \hspace{0.05em} \mathbb{R}^2\right)$$

If we set down,

$$ \Biggl \{ \begin{gather*} X = ln(x) \Longleftrightarrow x = e^{X}\\ Y = ln(y) \Longleftrightarrow y = e^{Y} \end{gather*} \qquad (5) $$

So,

$$ ln \left(\frac{x}{y} \right) = X - Y \longmapsto \frac{x}{y} = e^{X - Y} \qquad (8)$$

$$ \left(\mathbb{R}^2 \hspace{0.05em} \longmapsto \hspace{0.05em}\left[ \mathbb{R^*_+} \right]^2\right)$$

Both expressions \((5)\) and \((8)\) together show that:

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ \frac{e^a}{e^b} = e^{a - b}$$

Exponential to the power of n

Using the same logic as above,

$$ \forall n \ \mathbb{R}, \ x^n \longmapsto ln\left(x^n \right) = n \ ln(x) $$

$$ \left(\mathbb{R^*_+} \longmapsto \hspace{0.05em} \mathbb{R}\right)$$

If we set down,

$$ X = ln(x) \Longleftrightarrow x = e^{X} \qquad (5) $$

So,

$$ ln \left(x^n \right) = nX \longmapsto x^n = e^{nX} \qquad (9)$$

$$ \left(\mathbb{R} \hspace{0.05em} \longmapsto \hspace{0.05em} \mathbb{R^*_+}\right) $$

Both expressions \((5)\) and \((9)\) together show that:

$$ \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$ (e^a)^b = e^{ab} $$

Value of \(e\)

To determine an accurate value of the number \(e\), we can use the Taylor series of the exponential function at zero:

$$ \forall x \in \mathbb{R}, $$

$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ ... \ + \frac{x^n}{n!} + o(x^n)$$

$$ e^x = \sum_{k=0}^n \frac{x^k}{k!} + o(x^n)$$

Then, for \(x = 1\):

$$ e^1 = \sum_{k=0}^n \frac{1}{k!} + o(x^n) $$

This sum quickly converges towards \(2.7182818...etc\) (see table below).

$$ e \approx 2.7182818... $$

Link between powers to the base a or b and the exponential function\(: a^x, b^x, e^x\)

As well, we saw previously that:

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ log_a(x) = \frac{ln(x) }{ln(a)} $$

Now replacing \(x\) by \(a^x\), always strictly positive on \(\mathbb{R}\):

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall x \in \hspace{0.05em} \mathbb{R}, $$

$$ log_a(a^x) = \frac{ln(a^x) }{ln(a)} $$

$$ x \ ln(a) = ln(a^x) $$

Finally, applying the exponential function on each side,

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall x \in \hspace{0.05em} \mathbb{R}, $$

$$ a^x = e^{x \ ln(a)} $$

With this result, we can deduce of it that the derivative of \(a^x\). Indeed, in the chapter upon the determination of the origins of the logarithmic function, we had found that:

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, \ (a^x)' = \alpha \ a^x \qquad (2) $$

And in the aftermath, establishing a link between the logarithm of a base a and the natural logarithm we determined that: \(\alpha = ln(a)\).

So,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall x \in \hspace{0.05em} \mathbb{R}, $$

$$ (a^x)' = a^x \ ln(a) $$

In the general case, to find the relationship between two powers of x, we retrieve the previous formula:

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall b \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall x \in \hspace{0.05em} \mathbb{R}^*_+, $$

$$ log_a(x) = \frac{ log_b(x) }{log_b(a)} $$

Now replacing \(x\) by \(a^x\), always strictly positive on \(\mathbb{R}\):

$$ \forall a \in \hspace{0.05em} \Bigl[ \mathbb{R}^*_+ \hspace{0.2em} \backslash \{1\} \Bigr], \ \forall b \in \hspace{0.05em} \mathbb{R}^*_+, \ \forall x \in \hspace{0.05em} \mathbb{R}, $$

$$ log_a(a^x) = \frac{ log_b(a^x) }{log_b(a)} $$

$$ log_b(a^x) = x \ log_b(a) $$

Finally, applying \(b^x\) on each side:

$$ b^{log_b(a^x)} = b^{x \ log_b(a)} $$

$$ a^x = b^{x \ log_b(a)} $$

And finally,

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, \ \forall x \in \hspace{0.05em} \mathbb{R}, $$

$$ a^x = b^{x \ log_b(a)} $$

Finally, we can also notice that:

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ \Biggl \{ \begin{gather*} a^{ln(b)} = \left(e^{ln(a)} \right)^{ln(b)} = e^{ln(a)ln(b)} \\ b^{ln(a)} = \left(e^{ln(b)} \right)^{ln(a)} = e^{ln(a)ln(b)} \end{gather*}$$

So finally,

$$ \forall (a,b) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^2, $$

$$ a^{ln(b)} = b^{ln(a)} $$

But also for any logarithm:

$$ \forall (a,b, n) \in \hspace{0.05em} \left[ \mathbb{R}^*_+ \right]^3, $$

$$ \Biggl \{ \begin{gather*} a^{log_n(b)} \\ b^{log_n(a)} = \left(n^{log_n(b)} \right)^{log_n(a)} = n^{log_n(b)log_n(a)} = a^{log_n(b)} \end{gather*}$$

$$ a^{log_n(b)} = b^{log_n(a)}$$

Recap table of logarithm and exponential functions