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)?

  • $\begingroup$ Does this help? $\endgroup$ – Joseph Eck May 3 '18 at 4:28
  • $\begingroup$ @JosephEck ofcourse not did you even read the problem? $\endgroup$ – Turbo May 3 '18 at 4:29
  • $\begingroup$ 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 $E$? $\endgroup$ – Martín-Blas Pérez Pinilla May 3 '18 at 9:58
  • $\begingroup$ @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 $y>0$. $\endgroup$ – Turbo May 3 '18 at 10:48

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.