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)?
 A: 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: 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


*
  
*$\ln(xy) = \ln(x) + \ln(y)$
  
*$\frac{d}{dx} \ln(x) = \frac 1x$
  
*$\ln(x)$ is continuous. 
  
*$\ln(x)$ 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.
