Why is $\frac{f'(x)}{f(x)}$ always a constant?

Question

Today in class we learned that for exponential functions $f(x) = b^x$ and their derivatives $f'(x)$, the ratio is always constant for any $x$. For example for $f(x) = 2^x$ and its derivative $f'(x) = 2^x \cdot \ln 2$

$$\begin{array}{c | c | c | c} x & f(x) & f'(x) & \frac{f'(x)}{f(x)}\\ \hline -1 & \frac{1}{2} & 0.346 & 0.693 \\ 0 & 1 & 0.693 & 0.693 \\ 1 & 2 & 1.38 & 0.693\\ 2 & 4 & 2.77 &0.693& \end{array}$$

So as you can see, the ratio is the same and this is true for all functions of the form $b^x$ and its derivative. So my question is, why is the ratio always constant? Is there some proof or logic behind it or is just like that? Furthermore, what's the use of knowing this?

EDIT:

It seems that I have missed the fairly simple

$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{‌​b^x}}}=\ln\,b$$

But what's the use knowing and learning this? Will this reduce a step in the future or help solve a much harder problem more easily?

Answer 1

$$f=a^x$$ $$f'=\ln a \times a^x$$ So, $$f'=\ln a \times f$$ So, $$\dfrac{f'}{f}=\ln a= \text{constant}=0.693\Big|_{a=2}$$

Answer 2

$f'/f$ is called the Logarithmic Derivative of $f$.

By the Chain Rule $$ \frac{f'}{f}=\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x)) $$ Since $\log(b^x)=x\log(b)$ and $\frac{\mathrm{d}}{\mathrm{d}x}x\log(b)=\log(b)$, we get that $f'/f$ is constant.