Editing Exponential function (section)

===Equivalence proof===
For proving the equivalence of the above properties, one can proceed as follows.

The two first characterizations are equivalent, since, if {{tmath|1=b=e^k}} and {{tmath|1= k=\ln b}}, one has 
<math display=block>e^{kx}= (e^k)^x= b^x.</math>
The basic properties of the exponential function (derivative and functional equation) implies immediately the third and the last condition.

Suppose that the third condition is verified, and let {{tmath|k}} be the constant value of <math>f'(x)/f(x).</math> Since <math display = inline>\frac {\partial e^{kx}}{\partial x}=ke^{kx},</math> the [[quotient rule]] for derivation
implies that
<math display=block>\frac \partial{\partial x}\,\frac{f(x)}{e^{kx}}=0,</math> and thus that there is a constant {{tmath|a}} such that <math>f(x)=ae^{kx}.</math> 

If the last condition is verified, let <math display=inline>\varphi(d)=f(x+d)/f(x),</math> which is independent of {{tmath|x}}. Using {{tmath|1=\varphi (0)=1}}, one gets
<math display=block>\frac{f(x+d)-f(x)}{d} = f(x)\,\frac{\varphi(d)-\varphi(0)}{d}. </math>
Taking the limit when {{tmath|d}} tends to zero, one gets that the third condition is verified with {{tmath|1=k=\varphi'(0)}}. It follows therefore that {{tmath|1=f(x)= ae^{kx} }} for some {{tmath|a,}} and {{tmath|1=\varphi(d)= e^{kd}.}} As a byproduct, one gets that 
<math display=block>\left(\frac{f(x+d)}{f(x)}\right)^{1/d}=e^k</math>
is independent of both {{tmath|x}} and {{tmath|d}}.