# Is there an analytic proof of change of bases in logarithms?

Usually change of bases in logarithms is just observance

$$x=b^{\log_bx}\implies \log_kx={(\log_bx)}{(\log_kb)}\implies \log_kb=\frac{\log_kx}{\log_bx}.$$

Supposing apriori we do not know inverse of log is exponentiation but rather a function $E:\mathbb R\times\mathbb R\rightarrow\mathbb R$ such that $E(b,\log_bx)=x$ is there a proof that says $\log_kx=\log_k(E(b,\log_bx))={(\log_bx)}{(\log_kb)}$ (this essentially gets to how to show $\log_k(E(a,b))=b\log_ka$ without knowing $E$ is exponentiation)?

• Does this help? – Joseph Eck May 3 '18 at 4:28
• @JosephEck ofcourse not did you even read the problem? – Turbo May 3 '18 at 4:29
• If by definition $E(b,\log_b x) = x$, you are saying that $\log_b$ is the inverse of $y\longmapsto E(b,y)$. What another properties has $E$? – Martín-Blas Pérez Pinilla May 3 '18 at 9:58
• @Martín-BlasPérezPinilla assume it has same properties as exponential except it has two images everywhere. Assume $L(y,b)=L(-y,b)$ and that is $E(b,L(y,b))=|E(b,L(-y,b))|$ if $y>0$. – Turbo May 3 '18 at 10:48

## 2 Answers

Here is not exactly what you suggest but another approach which circumvents the fact that logarithm is the inverse of exponentiation. I list only the main points:

• start studying continuous functions satisfying the functional equation $L(xy) = L(x) + L(y)$
• derive properties like $L(x^r)=rL(x)$, $L(1) = 0$ etc.
• see that you may parametrize the functions $L$ by a parameter $b$ such that $$L_b(b) = 1$$
• see that $L$ is also differentiable $$\frac{L(x+h)-L(x)}{h}= \frac{1}{h}L\left(\frac{x+h}{h} \right) = L\left( (1+\frac{h}{x})^{\frac{1}{h}} \right)\stackrel{h\rightarrow 0}{\rightarrow} L\left( e^{\frac{1}{x}} \right) = \frac{1}{x}L(e)$$
• choose $b = e \rightarrow L_{e}'(x) = \frac{1}{x}$ and call it the natural logarithm $\ln x$.
• because of that and $L(1) = 0$ and writing the other logarithms as $\log_b x$, we get constants $c_b$ $$L_b(x) = c_b \int_1^x \frac{1}{t}dt = c_b\ln {x}$$
• it follows immediately $$\frac{\log_b(x)}{\log_k(x)} = \frac{L_b(x)}{L_k(x)}= \frac{c_b \ln x}{c_k \ln x} = \frac{c_b}{c_k}=\frac{\log_b e}{\log_k e}$$
• now substitute $x= k$ and note that $\log_k(k) = 1$: $$\log_b(k) = \frac{\log_b(x)}{\log_k(x)} = \frac{\log_b e}{\log_k e}$$

A long way around would be to start with the definition

\begin{align} \ln : (0, \infty) &\to \mathbb R \\ x&\mapsto \ln(x) = \int_0^x \frac 1t dt \end{align}

From this you can prove things like

1. $\ln(xy) = \ln(x) + \ln(y)$
2. $\frac{d}{dx} \ln(x) = \frac 1x$
3. $\ln(x)$ is continuous.
4. $\ln(x)$ is strictly increasing

and much more.

Because $y=\ln(x)$ is strictly increasing and continuous, it has a continuous inverse

\begin{align} \exp : \mathbb R &\to (0, \infty) \\ x&\mapsto \exp(x) \end{align}

We define $e = \exp(1)$ and we can use the usual methods to approximate its value.

We observe that, consequently, $\ln(e)=1$.

What if we wanted a function, say $\log_B$ that "behaves" like $\ln$ but has $\log_B(B)=1$?

Well, if we define $\log_B(x) = \dfrac{\ln(x)}{\ln(B)}$, then $\log_B(x)$ behaves very much exactly like $\ln(x)$ except that $\frac{d}{dx} \log_B(x) = \dfrac{1}{x \ln(B)}$. I will leave the details up to you.