Define L(x) to accumulate area under the hyperbola f(x) =  1 x , x>0. L( x ):= ∫ 1 x 1 t dt, x>0.   Then L( 1 )= ∫ 1 1 1 t dt =0. Define L(x) to accumulate area under the hyperbola f(x) =  1 x , x>0. L( x ):= ∫ 1 x 1 t dt, x>0.   Then L( 1 )= ∫ 1 1 1 t dt =0.

L( xy )= ∫ 1 xy 1 t dt = ∫ 1 x 1 t dt + ∫ x xy 1 t dt . Let u= t x  and du dt = 1 x , then L( xy )= ∫ 1 x 1 t dt + ∫ 1 y 1 ux xdu = ∫ 1 x 1 t dt + ∫ 1 y 1 u du =L( x )+L( y ). L( xy )= ∫ 1 xy 1 t dt = ∫ 1 x 1 t dt + ∫ x xy 1 t dt . Let u= t x  and du dt = 1 x , then L( xy )= ∫ 1 x 1 t dt + ∫ 1 y 1 ux xdu = ∫ 1 x 1 t dt + ∫ 1 y 1 u du =L( x )+L( y ).

L( x p )= ∫ 1 x p 1 t dt . Let  u p =t and p u p−1 du=dt, then L( x p )= ∫ 1 x p u p−1 u p du =p ∫ 1 x 1 u du =pL( x ). L( x p )= ∫ 1 x p 1 t dt . Let  u p =t and p u p−1 du=dt, then L( x p )= ∫ 1 x p u p−1 u p du =p ∫ 1 x 1 u du =pL( x ).

L( x y )=L( x y −1 )=L( x )+L( y −1 )=L( x )−L( y ). L( x y )=L( x y −1 )=L( x )+L( y −1 )=L( x )−L( y ).

Define e to be a number such that the area is 1: L( e ):=1= ∫ 1 e 1 t dt . Define e to be a number such that the area is 1: L( e ):=1= ∫ 1 e 1 t dt .

L( e L( x ) )=L( x )L( e )=L( x ) ⇒  e L( x ) =x ⇒  e L( 1 ) =1 ⇒  e 0 =1 L( e L( x ) )=L( x )L( e )=L( x ) ⇒  e L( x ) =x ⇒  e L( 1 ) =1 ⇒  e 0 =1

L has the properties of logarithms with a base of e. It was once called the Hyperbolic Logarithm, but now we call it the Natural Logarithm and rename it ln(x).  L has the properties of logarithms with a base of e. It was once called the Hyperbolic Logarithm, but now we call it the Natural Logarithm and rename it ln(x). 

We note in passing that with i:= −1  and  e iπ =cosπ+i sinπ=−1, then ln( e iπ )=i πln( e )=i π=ln(−1). We note in passing that with i:= −1  and  e iπ =cosπ+i sinπ=−1, then ln( e iπ )=i πln( e )=i π=ln(−1).

See Complex logarithm.

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This work is licensed by Stephen Kent Stephenson under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.

Last edited by Stephen Kent Stephenson on May 6, 2012 2:39 pm -0500.