Editing Natural logarithm (section)

==Definitions==
The natural logarithm can be defined in several equivalent ways.

===Inverse of exponential===
The most general definition is as the inverse function of <math>e^x</math>, so that <math>e^{\ln(x)} = x</math>. Because <math>e^x</math> is positive and invertible for any real input <math>x</math>, this definition of <math>\ln(x)</math> is well defined for any positive {{mvar|x}}.

===Integral definition===
[[File:Log-pole-x 1.svg|thumb|{{math|ln ''a''}} as the area of the shaded region under the curve {{math|1=''f''(''x'') = 1/''x''}} from {{math|1}} to {{mvar|a}}. If {{mvar|a}} is less than {{math|1}}, the area taken to be negative.]]
[[File:Log.gif|The area under the hyperbola satisfies the logarithm rule.  Here {{math|''A''(''s'',''t'')}} denotes the area under the hyperbola between {{mvar|s}} and {{mvar|t}}.|right|thumb]]

The natural logarithm of a positive, real number {{mvar|a}} may be defined as the [[area]] under the graph of the [[Hyperbola#Rectangular hyperbola|hyperbola]] with equation {{math|1=''y'' = 1/''x''}} between {{math|1=''x'' = 1}} and {{math|1=''x'' = ''a''}}.  This is the [[integral]]<ref name=":1" />
<math display="block">\ln a = \int_1^a \frac{1}{x}\,dx.</math>
If {{mvar|a}} is in <math>(0,1)</math>, then the region has [[negative area]], and the logarithm is negative.

This function is a logarithm because it satisfies the fundamental multiplicative property of a logarithm:<ref name=":2" />
<math display="block">\ln(ab) = \ln a + \ln b.</math>

This can be demonstrated by splitting the integral that defines {{math|ln ''ab''}} into two parts, and then making the [[Integration by substitution|variable substitution]] {{math|1=''x'' = ''at''}} (so {{math|1=''dx'' = ''a'' ''dt''}}) in the second part, as follows:
<math display="block">\begin{align}
   \ln ab = \int_1^{ab}\frac{1}{x} \, dx
   &=\int_1^a \frac{1}{x} \, dx + \int_a^{ab} \frac{1}{x} \, dx\\[5pt]
   &=\int_1^a \frac 1 x \, dx + \int_1^b \frac{1}{at} a\,dt\\[5pt]
   &=\int_1^a \frac 1 x \, dx + \int_1^b \frac{1}{t} \, dt\\[5pt]
   &= \ln a + \ln b.
 \end{align}</math>

In elementary terms, this is simply scaling by {{math|1/''a''}} in the horizontal direction and by {{mvar|a}} in the vertical direction.  Area does not change under this transformation, but the region between {{mvar|a}} and {{math|''ab''}} is reconfigured.  Because the function {{math|''a''/(''ax'')}} is equal to the function {{math|1/''x''}}, the resulting area is precisely {{math|ln ''b''}}.

The number {{mvar|e}} can then be defined to be the unique real number {{mvar|a}} such that {{math|1=ln ''a'' = 1}}.